Download Chapter 5 Trigonometric Functions

Chapter 7 Trigonometric Applications
Earth
Source: http://neo.jpl.nasa.gov/cgi-bin/db_shm?sstr=hermes&group=all
On Oct. 28, 1937, astronomer Karl Reinmuth of Heidelberg noticed an odd streak of light in a picture
he had just taken of the night sky. About as bright as a 9th magnitude star, it was an asteroid, close to
Earth and moving fast--so fast that he named it Hermes, the herald of Olympian gods. On Oct. 30,
1937, Hermes glided past Earth only 1.8 times as far away as the Moon (460,000 miles), racing
across the sky at a rate of 5 degrees per hour. Nowadays only meteors and Earth-orbiting satellites
move faster.
Hermes approaches Earth's orbit twice every 777 days. Usually our planet is far away when
the orbit crossing happens, but in 1937, 1942, 1954, 1974 and 1986, Hermes came harrowingly close
to Earth itself. The most recent close encounter happened on Nov. 4, 2003 when Hermes passed by
within 4,600,000 miles of earth. With the use of parametric equations we can look at the graphs of
the orbits of the earth and Hermes and tell whether we are in any danger of them colliding. Come
April, 23 2040 Hermes is supposed to pass quite close to earth again. With the parametric equations,
Hermes:




 (t  1.065)



1.01
x
 180 
 cos  3
7



1.9

10
(
t

1.065)




 180  
 1  0.624 cos  3
7

 1.9  10








  (t  1.065)
1.01
y
 180 
 sin  3
7



1.9

10
(
t

1.065)



 180   
 1  0.624 cos  3
7

 1.9 10


Earth:


0.999711
x
 cos(360(t  0.5438))
1

0.017
cos(360(
t

0.5438))




0.999711
y 
 sin(360(t  0.5438))
1

0.017
cos(360(
t

0.5438))


we can create graphs of the Earth and Hermes on our calculators and see how close they will come to
each other in the year 2040.
Chapter 7 Trigonometric Applications
Section 7.1 Right Triangle Definitions
Objectives

Understanding the right triangle definitions
We are now going to look at the third way of defining trigonometric functions. Here we will be using
the lengths of the sides of a right triangle to define each trigonometric function similar to what we
did in the first definition (coordinate definition) in section 5.3.
RIGHT TRIANGLE DEFINITIONS
Below is a right triangle with each of its sides labeled with respect to the angle θ.
Hypotenuse
Opposite
θ
Adjacent
The six trigonometric functions are defined as follows.
If θ is an acute angle of a right triangle, then we have the following definitions.
sin θ =
opposite
hypotenuse
cos θ =
adjacent
hypotenuse
tan θ =
opposite
adjacent
csc θ =
hypotenuse
opposite
sec θ =
hypotenuse
adjacent
cot θ =
adjacent
opposite
The words opposite, adjacent, and hypotenuse refer to the lengths of the sides of the right
triangle in relation to the acute angle θ.
This really isn’t anything new. When we talked about the definitions of the trigonometric functions
in section 5.3 we used x, and y and r, but those were just the sides of a right triangle that were
adjacent (x), opposite (y) and hypotenuse (r) to the angle θ. What is nice about this definition is that
some people find it easier to remember how to get trigonometric values from memorizing the words
(adjacent, opposite and hypotenuse) instead of the x, y and r, when working with application
problems. Let’s look at one application of this.
Discussion 1: Angle of Elevation (Forest Ranger)
Let’s figure out how high a tree is without having to climb it. The approach we are about to show
you is one a forest ranger might use. You would take a devise and find the angle between the
horizontal and the top of the tree. This, along with measuring how far you are from the trunk of the
tree, would allow you to calculate the height of the tree. Let’s say that you are standing 30 feet from
Chapter 7.1
the tree and the angle of elevation (the angle up from the horizontal to the line of sight to the top of
the tree) is 70°. Let’s figure out the tree’s height.
We know that we are 30 ft. from the tree
and the angle θ is 70°. Here is a sketch.
y
θ = 70°
30 ft
With respect to the angle θ we know its
(need better picture)
tan θ =
measurement and the side adjacent. We
opp
adj
want to find the length of the side opposite.
Thus tangent would be a good function to
use. Of course if you prefer more of a
tan 70° =
challenge use cotangent. Use the definition
of tangent and solve for the unknown
quantity (y).
y
30
30 (tan 70°) = y
(degree mode on TI)
82.4 ft = y
Let’s do another example.
Example 1: Trail Distance
You are on a hiking trip. You know that you started your hike at an altitude of 8,500 feet. The trail
took you to the top of a mountain that is 12,400 feet high. With the help of your GPS equipment you
calculate that you have moved horizontally on the earth 4 miles. Roughly speaking what was your
average angle of elevation?
Solution:
First we need to put all of our information
in common units. Let’s convert the miles to
 5280ft 
4 miles 
  21,120feet
 1mile 
feet.
Once again given this information tangent
12400 − 8500 = 3900 ft
would be our best choice to use on this
θ=?
problem.
21120 ft
(need better picture)
Solve for θ.
Substitute
We now need to take the inverse tangent of
both sides.
tan θ =
opp
adj
tan θ =
3900
21120
tan θ = 0.18466
θ = tan -1(tan θ) = tan -1 (0.18466)
Chapter 7 Trigonometric Applications
θ = 10.5°
(degree mode on TI)
Therefore on average you have been climbing a trail that has a 10.5 degree incline in elevation. We
only have one answer because this is a real problem where we are looking for an angle in just the
first quadrant.
Question 1: If you had been wearing a pedometer (a device that records how far you have walked)
while on your hike and it read 4.25 miles, would this roughly agree with your finding in example 1?
Another commonly used term besides angle of elevation is the term angle of depression. This is the
angle between the horizontal and something below the observer. Let’s look at an example.
Discussion 2: Canyon floor
If you have ever been to the Grand Canyon you might remember looking through telescopes that they
have at observation points that help you see key parts of the canyon. If you were looking through one
of these telescopes and noticed that when looking at the Colorado River below the angle formed
between the horizon and the line of sight to the river was 40°, how far would that part of the river be
from where you are knowing the fact that you are approximately 5000 feet above the river?
First draw a picture.
θ = 40
5000 ft
r=?
(need better picture)
Given this information sine would be our
best choice to use on this problem.
Solve for r.
sin θ =
opp
hyp
sin 40° =
5000
r
rsin 40° = 5000
r=
5000
5000

 7778.6feet
sin 40 0.64279
(degree mode on TI)
We are apparently 7778.6 feet from the river or roughly 1.5 miles. In this last example the angle θ
would be called the angle of depression since it was below the horizon.
Discussion 3: Bearings
There is a way to describe where something is by what is called bearings. A bearing is an angle that
is an acute angle which is measured from a north south line. For example if you are in Kansas City
the bearings to New York City would be north 77° east. In other words New York City is 77° to the
east of the line facing north out of Kansas City. For another example Seattle would be north 60° west
Chapter 7.1
from Kansas City. Las Angeles would be south 83° west of Kansas City and Miami would be south
46° east of Kansas City.
N
60° West
77° East
83° West
46° East
S
(picture of the U.S.A. with these five towns marked on it with north south and east west lines through
KC and the angles marked towards each of the other four towns like above)
For another example let’s look at a forest fire. With the help of trigonometry and the bearings from
two observation towers we can pinpoint the location of a forest fire. Let’s say that from one tower the
fire is north 30° west and from the other tower the fire is north 60° east and the two towers are 15
miles apart on an east west line. Here is a sketch of the situation.
fire
N
East
N
West
30°
60°
Tower two
Tower one
15 miles
To find out how far the fire is from each
tower we will use the fact that this has
d2
d1
formed a right triangle.
30°
60°
15 miles
We can now use sin 30° and the sin 60° to
calculate the distances d1 and d2.
sin 30° =
d1
15
sin 60° =
d2
15
Chapter 7 Trigonometric Applications
Solve for the two unknowns
15 sin 30° = d1
15 sin 60° = d2
7.5 miles = d1
13 miles  d2
Section Summary:
Answer Q1:
Yes.
3900 ft
tan(θ)=

4.25 m =
22440 ft
3900
22440
θ = 9.85°
If θ is an acute angle of a right triangle, then we have the following definitions.
sin θ =
opposite
hypotenuse
cos θ =
adjacent
hypotenuse
tan θ =
opposite
adjacent
csc θ =
hypotenuse
opposite
sec θ =
hypotenuse
adjacent
cot θ =
adjacent
opposite
The words opposite, adjacent, and hypotenuse refer to the lengths of the sides of the right
triangle in relation to the acute angle θ.



The angle of elevation is the angle up from the horizontal to the line of sight of an object above
the observer.
The angle of depression is the angle between the horizontal and the line of sight to something
below the observer.
A bearing is an angle that is an acute angle which is measured from a north south line.
Chapter 7.1
SECTION 7.1 PRACTICE SET
(14) For each right triangle find the values of the six trigonometric functions for Angle A:
1.
2.
20
33
3.
20
10
A
15
A
A
15
4.
18
24
A
(514)
B
a
c
C
A
b
ABC is a right triangle with the right angle at C. From the following given information find the six
trigonometric function values of the angle B.
5. a = 5; c = 10
6. b = 12; c = 13
7. a = 5; c = 10
8. b = 7; a = 7 3
9. a = 7; b = 7
10. b = 8; c = 8 2
11. b = 15; c = 22
12. a = 15; b = 19
13. a = 8; b = 9
14. a = 15; b = 19
(1520)
B
a
C
c
A
ABC is a right triangle with the right angle at C. Use your calculator (if necessary) to find the
missing values. (approximate to 4 decimal places)
15.  A = 50o; a = 8
B= b= c=
16.  B = 20o; b = 10
A= a= c=
17.  A = 43o; b = 12
B= a= c=
18.  B = 56o; a = 20
A= b= c=
19.  A = 58o; c = 30
B= a= b=
20.  B = 62o; c = 40
A= a= b=
Chapter 7 Trigonometric Applications
(2124) Angles of Elevation
21. The angle of elevation from ground level 50 feet away from a flag pole is 50 o. What is the
height of the flag pole?
NEED PICTURE
x
50o
50 feet
22. Standing 50 yards from the base of a cliff the angle of elevation from ground level to the top
of the cliff is 80o. How high is it from the bottom to the top of the cliff?
NEED PICTURE
x
80o
50 yards
23. It is 1,800 feet to the top of a mountain from the base of the mountain. You traveled
an inclined path of 3,200 feet to get to the top of the mountain. What is the average angle
of elevation for the path?
NEED PICTURE
1,800
feet
3,200
feet
x
24. You observe a plane that is flying at an altitude of 10,000 feet. A camera is taking picture
from ground level at a point 1,000 feet away from the point the plane is flying directly over.
What is the angle of elevation of the camera?
Plane
NEED PICTURE
10,000
feet
x
1,000
feet
Camera
Chapter 7.1
(2530) Angle of Depression
25. You are laying down at the edge of a cliff 800 feet high, looking at a river that is 300 yards
from the base of the cliff. Assuming the cliff is perpendicular to the ground, what is the angle
of depression for your line of sight?
NEED PICTURE
Person
x
800
feet
300 yards
26. You are in a balloon 1,000 feet above the ground. You observe your car on the ground
200 yards from the base of the balloon. What is the angle of depression for you line of sight?
NEED PICTURE
Balloon
x
1,000
feet
Car
200 yards
27. A 100 foot flagpole casts a shadow 53 feet long. What is the angle of depression of the sun?
NEED PICURE
x
100
feet
50
feet
Chapter 7 Trigonometric Applications
28. Just before you go over a waterfall that is 80 feet high you observe a boat 50 feet from the
base of the waterfall. What is the angle of depression of your line of sight?
NEED PICTURE
x
80
feet
50
feet
Boat
29. A kite is flying at a height of 90 feet and the angle of depression between the kite and the place
the string is attached at ground level is 33o. How much string is being used?
NEED PICTURE
Kite
90
feet
x
33o
30. A person vacationing in Mexico climbs 200 feet up the side of a pyramid. The angle of
depression is 55.3o. How tall is the pyramid?
NEED A PICTURE
x
55o
200
feet
x
(3134) Bearings
31. There are two observations towers 20 miles apart on an East-West line. From one tower
the forest fire is S 35o E and from the other tower the forest fire is S 50 o W. How far is the
fire from each tower?
NEED PICTURE
N
TOWER
20 miles
35o
S
N
TOWER
50o
FIRE
S
Chapter 7.1
32. Two ships at sea are 20 miles apart from each other on an East-West line. One ship’s bearing
from a lighthouse is N 20o W and the other ship’s bearing from the light house is N 30 o E. How
far from the light house is each ship?
NEED PICTURE
N
SHIP
20 miles
30o
S
N
SHIP
20o
LIGHTHOUSE
S
33. A boat travels on a course bearing N 41.3 o W for 38 miles. How many miles north and how
many miles west has the boat traveled?
34. A boat travels on a course bearing S 38.7 o E for 52 miles. How many miles south and how
many miles east has the boat traveled?
Chapter 7 Trigonometric Applications
Section 7.2 Law of Sines
Objectives

Understanding the law of sine

Understanding the Ambiguous Case

Understanding the area of any triangle
In this section we are going to talk about how to find angles and sides of triangles that are not right
triangles.
Oblique Triangle  An oblique triangle is one that is not a right triangle.
Generally we tend to label the sides and angles of triangles in the following way.
A
b
c
B
a
C
The angles are labeled with capital letters and the sides are labeled with lower case letters that
correspond to the angles opposite of them. As you go around a triangle you move from an angle to a
side to an angle to a side and so forth. In geometry we use the following notation to signify which
pieces of a triangle are known quantities.
SAS  Stands for side-angle-side which tells us that we know the values of two sides and
the angle between them.
ASA − Stands for angle-side-angle which tells us that we know the values of two angles
and the side in between them.
SSS − Stands for side-side-side which tells us that we know the values of all three sides.
It turns out that if you know any of the above three mentioned situations, SAS, ASA, and SSS, then
you can find the values of all of the other parts of the triangle and that there will be only one unique
answer. For a counterexample, if you only know the values of the three angles of a triangle (AAA),
then there would be many answers for the sides of the triangle. There are an infinite number of
triangles with the same angles and all of them would be similar to each other.
It is true that if you know three of the measurements of a triangle and at least one of those is
a side, then you can find the measurements of the other parts of the triangle, but it may turn out that
the triangle isn’t possible or that there are two possibilities. Let’s begin by discovering a way to solve
all of these different triangles.
Chapter 7.1
THE LAW OF SINES
Discussion 1: The Law of Sines
Let’s look at how we arrive at the law of sines. We begin with an oblique triangle. It doesn’t matter
which type of oblique triangle we have the law of sines can be proven to be true for them all. So,
let’s just take one example and work with it.
A
b
c
h
D
We begin by creating two right triangles
sin C 
h
b

h  b sin C
sin B 
h
c

h  c sin B
(ADC, and ADB) as seen above. We know
from previous work that sin  
opp
.
hyp
C
a
B
Next use substitution to get,
b sin C  c sin B
Lastly divide both sides of the equation by
b sin C c sin B

bc
bc
bc.
(set h’s equal)
sin C sin B

c
b
By creating right triangles from vertices B and C one can discover all the relationships that exist in
the law of sines.
The Law of Sines  In any triangle ABC with sides a, b, and c, the following is always
A
true.
sin A sin B sin C


a
b
c
b
or
a
b
c


sin A sin B sin C
C
c
a
B
Example 1: Finding the Measurements of a Triangle
Given a triangle ABC with measurements, A = 27°, B = 82°, and a = 12 inches, what are the
other measurements of the triangle?
Solution:
C = ?
This is an AAS triangle which means
b=?
we can find the third angle, so we
a = 12
could also view this as an ASA
triangle. There will only be one
A = 27°
possible triangle.
Sum of angles of triangles equal 180°.
27° + 82° + C =180°
c=?
B = 82°
Chapter 7 Trigonometric Applications
C = 71°
Use law of sines to find the other two
sides.
a
b
c


sin A sin B sin C
12
b
c


sin 27 sin 82 sin 71
C = 71°
b = 26.175
12
b

sin 27 sin 82
a = 12
sin82
A = 27°
c = 24.99
B = 82°
12
c

sin 27 sin 71
12
b
sin 27
sin 71
26.175 inches  b
12
c
sin 27
24.99 inches  c
Let’s do another example similar to discussion 3 in section 7.1.
Example 2: Forest Firers
Two observation towers are 45 miles apart. The bearings from tower (A) to tower (B) is north 24°
west. A fire is spotted from the two towers. The bearings from tower A is north 18° east and the
bearings from tower B is north 48° east. How far is the fire from each tower?
Solution:
First we should draw and label a
Fire C
picture of our problem.
a=?
48°
Tower B
b=?
18°
d
24°
45 miles
Tower A
Angle A would be 18° plus 24° which is 42°. Angle B is a little harder to find. We need to
know what angle d is first. But from section 5.1 we know from Truth 1 that alternate
interior angles are equivalent. So d = 24°. Therefore B = 180° − 48° − 24° = 108°. Now
we know two angles and a side, we should be able to answer the question.
First find the third angle measurement.
Fire angle = 180° − 108° − 42° = 30°
Now, use the law of sines to find the
c
b
a


sin C sin B sin A
two sides.
45
b
a


sin 30 sin108 sin 42
Chapter 7.1
Fire C
60.22 miles
Tower B
108°
45 miles
30°
45
b

sin 30 sin108
sin108
85.6 miles
45
a

sin 30 sin 42
45
b
sin 30
sin 42
85.6 miles  b
45
a
sin 30
60.22 miles  a
42°
Tower A
The fire is now located and fire fighters will be able to determine what would be the best forest roads
to use to get to the fire the quickest.
THE AMBIGUOUS CASE
Unfortunately life isn’t always simple. If you know two sides and an angle opposite one of the sides
(SSA triangle) then it may be that there isn’t a triangle that could exist with those measurements or
there is only one that could or there are two that could. It is easiest to show you some pictures of
what can happen with the ambiguous case. We will assume that in each picture that we know the
measurements of angle A and sides a and b.
The known angle A is an acute angle.
Situation 1
The side opposite the known angle is to
a
b
short. A triangle isn’t formed.
A
Situation 2
The side opposite the known angle is just
b
barely long enough. A right triangle is
formed.
a
A
Situation 3
The side opposite the known angle is a little
b
longer than the height of the triangle but is
a
less than the other side. Two triangles can
a
A
be formed.
Situation 4
The side opposite the known angle is longer
b
than the other known side. One triangle is
formed.
A
a
Chapter 7 Trigonometric Applications
The known angle A is an obtuse angle.
Situation 5
a
The side opposite the known angle is to
short. A triangle isn’t formed.
b
A
Situation 5
The side opposite the known angle is longer
a
than the other known side. One triangle is
b
A
formed.
Discussion 2: No Triangle Situation
Let’s look at how we approach finding all of the unknown measurements of the following triangle.
A = 38°, b = 13 and a = 7.
This is an SSA triangle. The question is, does
a reach far enough to form one or more
b = 13
A = 38°
triangles.
Let’s begin by using the law of sines. Use this
form since we are looking for an angle (B).
Substitute and solve for angle B.
a=7
sin B sin A

b
a
sin B sin 38

13
7
sin B  13
sin 38
 1.14
7
Since sin B is equal to a value greater than 1 which is impossible, a must not be long
enough to form a triangle.
Example 3: Two Triangles
Find all the measurements of the following triangle.
A = 29°, b = 7 and a = 4.
Solution:
This is an SSA triangle. The question is, does
a reach far enough to form one or more
b=7
A = 29°
triangles.
Let’s begin by using the law of sines. Use this
form since we are looking for an angle (B).
Substitute and solve for angle B.
sin B sin A

b
a
sin B sin 29

7
4
a=4
Chapter 7.1
sin B  7
sin 29
 0.8484
4
B = sin 1(0.8484)  58
With B = 58°, angle C = 93° (the third angle
would be 180° − 29° − 58° = 93°) so we can
find side c.
c
a

sin C sin A
c
4

sin 93 sin 29
c  sin 93
4
 8.24
sin 29
C = 93°
1st triangle.
b=7
a=4
A = 29°
c = 8.24
B = 58°
Since sine is positive in the first and second
180° − 58° = 122°
quadrants and we found a value for B to be
This is a second quadrant angle with the
58° it is possible that there is another value for
same reference angle as a 58° angle.
angle B which would be in the second
Therefore sin 58° = sin 122°.
quadrant where the sine function would also
yield an output of 0.8484.
We have a possible 2nd triangle. With B =
If angle B and C are the same (isosceles
122° (the third angle C would be 180° − 29° −
triangle) then side a would equal side c.
122° = 29°). Notice that this would make our
triangle be an isosceles triangle.
C = 29°
2nd triangle.
b=7
A = 29°
c=4
a=4
B = 122°
Question 1: If the sides in the last example were such that angle B had turned out to be 25°, would
there be an easy way of knowing whether there were two triangles possible or just one?
Let’s summarize what will happen algebraically and what it will mean to you as you try solving for
sides and angles of triangles.
If you know sides a and b and angle A of triangle ABC, then you can find the other parts of
the triangle by using the law of signs and you will encounter one of these three cases.

If while in the process of solving for sin B you find it to be equal to a value larger
than one, then this means that there does not exist a triangle with these known
Chapter 7 Trigonometric Applications
values. (like in discussion 2)

If while in the process of solving for sin B you find it to be equal to one, then this
means that there exist a right triangle where angle B is a right angle.

If while in the process of solving for sin B you find it to be equal to a value that is
less than one, then this means that there exist one or two triangles. (ambiguous
case)
1.
The first triangle results from the value for angle B from the first quadrant.
2.
The second triangle, if it exists, comes from the value for angle B in the
second quadrant but you must check to see if it is possible.
Let’s look at one more example where we get one triangle but not two.
Example 4: Only One Triangle
Find all the measurements of the following triangle.
A = 37°, b = 3 and a = 5.
Solution:
Here is a sketch of our triangle.
a=5
b=3
A = 37°
We will once again use the law of sines to
assist us in finding the other measurements
sin B sin A

b
a
of this triangle.
Substitute and solve for angle B.
sin B sin 37

3
5
sin B  3
sin 37
 0.3611
5
B = sin 1(0.3611)  21.17
With B = 21.17°, angle C = 121.83° (the
third angle would be 180° − 37° − 21.17° =
121.83°) so we can find side c.
c
a

sin C sin A
c
5

sin121.83 sin 37
c  sin121.83
5
 3.09
sin 37
C = 121.83°
1st triangle.
b=3
a=5
A = 37°
c = 3.09
B = 21.17°
Since sine is positive in the first and second
180° − 21.17° = 158.83°
quadrants and we found a value for B to
This is a second quadrant angle with the
be 21.17° it is possible that there is another
same reference angle as a 21.17° angle.
Chapter 7.1
value for angle B which would be in the
Therefore sin 21.17° = sin 158.83°.
second quadrant where the sine function
would also yield an output of 0.3611.
We do not have a possible 2nd triangle. We don’t have a second triangle because the second
value for angle B = 158.83° when added to angle A, puts us over the 180° limit for the sum
of the angles of a triangle (158.83° + 37° = 195.83°).
Question 2: Look at what happened in example 3 and 4 with the first quadrant angle’s value we
found for angle B. Do you see an easy way of telling if you are going to have one or two triangle?
THE AREA OF ANY TRIANGLE
1
2
As you probably already know the area of a triangle is one half the base times the height ( A  bh ).
In order to use this formula you need to know the height of the triangle which may not be given to
you. You have used this formula the most in your past when working with right triangles. But now
let’s look at just any old triangle in general and what type of formula that might produce.
Discussion 3: Find Area in General
With a general triangle let’s see what it will take to find a formula
C
b
for its area. There are a couple of different shaped triangles that we
A
would need to look at to prove the formulas to come, but we think
h
c
that it suffices to show you just one example here.
If you wanted to find the area of this
sin A 
h
b

h  b sin A
sin B 
h
a

h  a sin B
triangle it would be imperative that you
figure out the value of h. Since h forms two
right triangles we can do this easily many
different ways.
So, in this example to find the area we
A
would use one of these two choices.
If c were not the base and we changed this
sin C 
triangle’s orientation so that a or b were the
base then we would see this as a possibility
for h.
1
1
1
bh  c  b sin A   c  a sin B 
2
2
2
A
h
a

h  a sin C
1
1
1
bh  a  b sin C   b  a sin C 
2
2
2
From this discussion we get the following statement.
In any triangle ABC the area (A) is found using one of the following:
a
B
Answer Q1:
Yes. If angle B
had been 25°
then the
supplementary
angle would
be 155°. So the
two possible
angles for B
would be 25°
and 155°. With
the 25° angle
C would be
180-29-25
=126°. With
the 155° angle
C would be
180-29-155
=−4°. This is
impossible.
Thus there is
only one angle
possible.
Chapter 7 Trigonometric Applications
1
2
A  ac sin B ,
1
2
1
2
A  ab sin C ,
A  bc sin A
Notice that whichever two sides are used in the formula the angle used is always the angle
opposite the third side.
Example 5: Finding Area
What is the area of a triangle whose angle B = 87° and where side a = 13 and side c = 4?
Solution:
The formula we would use given this info is
A  ac sin B
Substitute in the values we know and solve
A
for A.
1
2
1
13 4  sin 87  25.96 sq units
2
Section Summary:

Oblique Triangle  An oblique triangle is one that is not a right triangle.

The Law of Sines  In any triangle ABC with sides a, b, and c, the following is always true.
sin A sin B sin C


a
b
c

a
b
c


sin A sin B sin C
or
If you know two sides and an angle opposite one of the sides (SSA triangle) then it may be
that there isn’t a triangle or there is only one or there are two possible. If you can’t find and
angle ( sin B > 1) then you have no triangle formed. If you find an angle that is smaller than
the one known then you have only one triangle and if you find an angle that is larger than
the one known then you have two triangles possible.

In any triangle ABC the area (A) is found using one of the following:
1
2
A  ac sin B ,
1
2
A  ab sin C ,
1
2
A  bc sin A
Notice that whichever two sides are used in the formula the angle used is always the angle
opposite the third side.
Chapter 7.1
SECTION 7.2 PRACTICE SET
A
c
b
B
a
C
Use the above figure for problems (1−22)
(1−8) Determine the length of the remaining sides and the measurement of the remaining angles for
each of the following.
1. A= 50o; B= 102o; c = 5 feet
2. C= 48o; B= 98o; = 10 feet
3. A= 48.3o; C= 39.5o; a= 15.2 cm
4. B= 103.5o; C= 43.2o; c= 14.3 cm
5. A= 32o; B= 98o; c= 10 inches
6. B= 102o; C= 28o; a= 12 inches
7. A= 32.5o; C= 28.3o; a= 32.3 miles
8. B= 101.9o; C= 105.2o; a= 10.9 miles
(9−22) Same instructions as (1−8) (Ambiguous Cases)
9. a = 35 cm; b = 25 cm;  A= 33o
10. a = 30 inches; b = 53 inches;  B= 72o
11. a = 10 feet; c = 8 feet;  A= 50o
12. b = 12 inches; c = 15 inches;  C= 63o
13. a = 8 m; c = 5 m;  C= 35o
14. b = 12 cm; c = 18 cm;  B= 62o
15. a = 12 feet; b = 18 feet;  A= 53o
16. b = 15 inches; c = 12 inches;  C= 43o
17. a = 9 cm; b = 8 cm;  B= 98o
18. b = 11 inches; c = 8 inches;  C= 101o
19. a =8 inches; b = 5 inches;  C= 83o
20. b =9 feet; c = 6 feet;  C= 72o
21. a = 10 cm; c = 7 cm;  C= 63o
22. a = 3 m; b =5 m;  A= 79o
(23−26) Find the area of the following:
C
b
A
a
c
B
23. a = 6 inches; c = 8 inches;  B= 71o
24. a = 5 cm; b = 12 cm;  C= 69o
25. b = 0.83m; c = 0.95m;  A= 62.5o
26. a = 3.95 cm; c = 5.83 cm;  B=43.8o
Answer Q2:
If angle B is
smaller than
angle A a
second
possibility will
never exist.
Chapter 7 Trigonometric Applications
27. A surveyor takes two sightings 1,000 feet apart in a direct line to the top of a mountain.
The first angle of elevation is 53o and the second angle of elevation is 33o. What is the
height of the mountain?
NEED PICTURE
x
(x should labeled mountain)
53o
33o
1,000 feet
28. A tree trimmer charges $30 a foot to cut down a tree. To find the height of a tree he measures
the angle of elevation between the tree and two place 101 feet apart. The first angle of elevation
is 53o and the second angle of elevation if 43o. How tall is the tree and what is the charge for
cutting down the tree?
NEED A PICTURE
x
(x should be labeled tree)
53o
43o
101 feet
29. Two lighthouses 30 miles apart observe a ship sailing the ocean. The bearing from
lighthouse A to the ship is N 50o E and the bearing from lighthouse B to the ship is
N 43o W. How far is each lighthouse from the ship?
NEED PICTURE
SHIP
50o
A
43o
40o
47o
B
30. Two ships A and B 20 miles due east-west of each other receive a SOS sign from a ship. The
bearing from ship A to the ship sending the SOS is S 53o E and the bearing from ship B to the
ship sending the SOS is S 47o W. How far is the ship sending the SOS signal from Ship A and
how far from ship B.
NEED A PICTURE
SHIP A
20 miles
SHIP B
53O
470
SHIP
Chapter 7.1
31. An airplane is flying between two observation stations due east-west of each other are 20
miles apart. The angle of elevation for one observation station is 83 o and the angle of
elevation for the other station is 74o. How far is the airplane from each observation station?
NEED PICTURE
AIRPLANE
83O
74O
O.S.
O.S.
32. One person is standing on one side of a flagpole and the angle of elevation is 50 o and a second
is standing 100 feet due east of the first person on the other side of the flagpole. The angle of
elevation from the second person is 62o. How far from the top of the flagpole is each person
standing?
TOP OF THE FLAGPOLE
NEED PICTURE
62O
50O
100 feet
33. From problem (31) how high is the plane flying?
34. From problem (32) what is the height of the flagpole?
35. A surveyor wants to know the distance from a power plant to a factory across the river from the
power plant. The surveyor find the angle for the line of sight from the power plane to the factory
is 115o10. The surveyor the picks a spot 360 meters down the river on the side of the power
plant and finds the angle for the line of sight from that point to the factory is 18 010. How far is
it from the power plant to the factory?
NEED PICTURE
POWER PLANT 360 meters
POINT DOWN STREAM
115o20
18o10
FACTORY
Chapter 7 Trigonometric Applications
36. Two hikers A and B 100 yards from each other are looking at the base of a tree. Hiker A has a
line of sight of 58o10 and hiker B has a line of sight of 103o20. How far from the base of the
tree is each hiker?
NEED PICTURE
A
100 yards
B
50o10
103020
TREE
37. A ship is sailing due north. At a certain time the bearing from the ship to a lighthouse
12.5 km away is N 35o E. After 1 hour of sailing due north the bearing from the ship
to the light house is S 43o E.
NEED PICTURE
SHIP ONE HOUR LATER
43o
LIGHTHOUSE
12.5 km
35o
SHIP AT BEGINNING
a. How far has the ship traveled in the one hour period?
b. How far is the ship from the lighthouse one hour later?
c. What is the average speed of the ship?
38. An automobile traveling due east on a road. At the time the automobile is 12 miles from a town
the bearing from the automobile to the town is N 58 o W. Thirty minutes later the bearing from
the automobile to the town is N 59o E.
NEED PICTURE
TOWN
12 miles
58o
59o
CAR AT BEGINNING
CAR AFTER 30 MIN
a. How far did the automobile travel in 30 minutes?
b. How far was the automobile from the town after 30 minutes?
c. What was the average speed of the automobile, in miles per hour, for the 30 minutes?
Chapter 7.1
Section 7.3 Law of Cosines
Objectives

Understanding the law of cosine
Unfortunately the law of sines doesn’t solve every triangle for us. In this section we are going to
learn about the law of cosines. This law is a little more involved but necessary for certain types of
problems. With the law of sines, as long as you knew three pieces of information about the triangle
where, two of those were, a side and the angle opposite of it, you could solve for all the parts of the
triangle if it existed. If, on the other hand, you don’t know a side and the angle opposite it then, we
will need to use the law of cosines.
THE LAW OF COSINE
Let’s start off as we did with the law of sines and show you how we arrive at the formula.
A(x, y)
y
c
D
x
b
B(0, 0)
a
C(a, 0)
We, as before, will show you one example of deriving the formula.
In the triangle above we know the
sin B 
y
c

y  c sin B
cos B 
x
c

x  c cos B
following.
Thus, the point A has the following

A(x, y)
A(c cos B, c sin B)
coordinates
Now we use the distance formula.
b
 c cos B  a 2   c sin B  0 2
Square both sides
b 2   c cos B  a    c sin B  0 
Clear parentheses
b2  c2 cos2 B  2ac cos B  a2  c2 sin 2 B
Collect the c2 terms
b2  c2 cos2 B  sin 2 B  2ac cos B  a 2
Pythagorean identity
b2  c 2  a 2  2ac cos B
2


2
Chapter 7 Trigonometric Applications
Depending on the orientation of our triangle we get three different formulas.
Law of Cosines − In any triangle ABC The following is true.
C
a 2  b2  c 2  2bc cos A
b
b2  a 2  c2  2ac cos B
A
c2  a 2  b2  2ab cos C
a
c
B
Notice that whichever side is on the left side of the formula the angle opposite that side is the one
you cosine. Let’s do an example.
Example 1: Using Law of Cosines
Given triangle ABC and the values a = 4, c = 6 and B = 43°, what are the other parts of triangle
ABC?
Solution:
Since we do not know a side and its opposite
b2  a 2  c2  2ac cos B
angle we need to use the law of cosines. The
formula we need is the one that has b on the
left side of the formula.
Substitute in the values given and solve for b.
b2  42  62  2  4  6  cos 43
b2  16  36  48  0.731
b2  16.9
b  4.1
Now we could use the law of sines to find
the measurement of the angle A.
sin B sin A

b
a
sin 43 sin A

4.1
4
sin A  4
sin 43
 0.6654
4.1
A = sin 1  0.6654   41.7
Now find angle C from the fact that the sum
C = 180° − 41.7° − 43° = 95.3°
of the angles of a triangle equals 180°.
This problem worked out the best when we use a lot of what we have learned: law of cosines, law of
sines, and the sum of the angles of a triangle equal 180°.
Let’s look at another example of where we need to use the law of cosines.
Discussion 1: Three sides known
Chapter 7.1
Given triangle ABC and the values a = 11, b = 5 and c = 14, let’s find the measurements of the three
angles of triangle ABC.
Since we do not know a side and its opposite
a 2  b2  c 2  2bc cos A
angle we need to use the law of cosines.
112  52  142  2  514cos A
Substitute in the values given and solve for cos
A, then solve for A.
121  25  196  140cos A
140cos A  25  196  121
If cos 1 x had turned out to be negative then we
cos A 
would have known that this was an angle from
100
 0.7143
140
A = cos1  0.7143  44.4
the second quadrant. Remember cosine inverse
is only defined for the first and second quadrant
angles.
If we had solved for angle C, here is what we
142  112  52  2 11 5cosC
would have seen.
cos C 
50
 0.4545
110
cos1  0.4545  117
Now we can use the law of sines to find the
measurement of angle C.
sin C sin 44.4

14
11
sin C  14
 sin C sin A 
 c  a 


sin 44.4
 0.89
11
C = sin 1  0.89   62.9
If we used this value for C we would end up being incorrect. If we believe that angle C =
62.9°, then that would mean that 180° − 62.9° − 44.4° = 72.7° = B. But b is the shortest
side which means that angle B must be the smallest angle which it isn’t by these
calculations.
A good rule of thumb, when using the law of cosines to find an angle, would be to find
the biggest angle first. (The biggest angle will always be opposite the longest side.)
This may not always be possible because of the information given to you in a problem. Be
very careful when solving a problem if this is the case.
If we had found angle C first we would see that
it is 117°.
142  112  52  2 11 5cosC
cos C 
50
 0.4545
110
cos1  0.4545  117
sin B sin C

b
c
sin B sin117

5
14
sin B  5
sin117
 0.3182
14
B = sin 1  0.3182  18.6
Chapter 7 Trigonometric Applications
The last angle.
A =180° − 18.6° − 117° = 44.4°
Let’s do another example.
Example 2: SAS Triangle
Given triangle ABC and the values A = 12.9°, b = 9 and c = 11, what are the measurements of the
other parts of triangle ABC?
Solution:
Since we do not know a side and its opposite
a 2  b2  c 2  2bc cos A
angle we need to use the law of cosines.
Substitute in the values given and solve for a.
a 2  92  112  2  9 11 cos12.9
a 2  81  121  198  0.9748
a 2  8.99
a3
Now we could use the law of sines to find
the measurement of the angle C.
sin A sin C

a
c
sin12.9 sin C

3
11
sin C  11
sin12.9
 0.8186
3
C = sin 1  0.8186   54.9
Question 1: Do you think that this is the correct answer for angle C?
Another possible value for angle C.
C = 180° − 54.9° = 125.1°
This means that angle B must be this amount.
B = 180° − 125.1° − 12.9° = 42°
Notice that we needed the bigger value for angle C since it was opposite the largest side. If we had
used the smaller amount then angle B would have turned out to be bigger than angle C which
couldn’t be correct. Let’s look at an application of the law of cosines.
Example 3: Surveying
A surveying team wants to calculate the current length of a lake. They set up their equipment as
shown in the figure and have made the measurements shown.
Chapter 7.1
Surveyor C
Surveyor A
832 yds
72°
1260 yds
Surveyor B
What is the length of the lake?
Solution:
Since we do not know a side and its opposite
b2  a 2  c2  2ac cos B
angle we need to use the law of cosines.
Substitute in the values given and solve for b.
b2  8322  12602  2 8321260 cos72
b 2  692224  1587600  2096640(0.309)
b2  1631926.6
b  1277.5yards
The lake is currently 1,277.5 yards across.
We now know the length across the lake which could be compared to previous measurements and
used by many different people in their study of the environment.
Section Summary:
Chapter 7 Trigonometric Applications
SECTION 7.3 PRACTICE SET
B
c
A
a
b
C
(1−14) Determine the length of any remaining sides and the measurement of any remaining angles
for each of the following.
1. a = 5 inches; b = 7 inches; c = 8 inches
2. a = 9 feet; b = 7 feet; c = 13 feet
3. a = 3.2 cm; b = 5.7 cm; c = 4.9 cm
4. a = 8.3 m; b = 9.4 m; c = 15.3 m
5. a = 9 yards; b = 7 yards; m C= 58o
6. b = 15 m; c = 11 m; m A= 38o
7. a = 5.1 feet; b = 10.3 feet; c = feet
8. a = 15.2 cm; b = 40.3 cm; c = 26.1 cm
9. a = 3.8 inches; c = 9.2 inches; m B= 58.3o
10. a = 4.9 yards; b = 8.3 yards; m C= 78.2o
11. a = 38.3 m; c =42.5 m; m B= 103.8o
12. b = 53.2 cm; c = 49.3 cm; m A= 115.35o
13. a = 3.2 feet; b = 8.74 feet; m C=123o30
14. b = 2.83 m; c = 9.54 m; m A= 118o20
15. A surveyor wants to know the distance from house A to house B that has a lake between them.
From a point C the distance to house A is 130 yards and the distance from point C to house B
is 210 yards. The angle between the two measurements is 68 o. What is the distance between the
two houses?
NEED A PICTURE
B
Answer Q1:
If it is correct
then angle B
would be
180-12.9-54.9
=112.2°. But if
LAKE
side c is larger
A
than side b,
then angle C
130 yards
should be
68o
larger than
C
angle B, which
it isn’t. Thus
this must not 16. From problem (15), house A is 320 yards from point C and house B is 210 yards from point C
be the correct
and the angle between is 53o. What is the distance between the two houses?
answer.
17. Two ships leave port at the same time, traveling on a course that has an angle of 128 o between
them. One of the ships has traveled 330 miles and the other 280 miles. How far apart are the
ships?
NEED A PICTURE
280 miles
128o
330 miles
Chapter 7.1
18. Two airplanes are flying at the same height. The two planes are flying so the angle
between them and their collision point is 1090. One plane is 380 miles from the
collision point and the other plane is 470 miles from the collision point. How far
apart are the planes?
NEED PICTURE
PLANE
PLANE
380 miles
470 miles
109o
19. A plane flies from city A a distance of 300 miles due north , then turns through an angle of 40 o
and flies to city C a distance of 250 miles.
NEED PICTURE
CITY C
250 miles
40o
300 miles
CITY A
a. How far is it from city A to city C?
b. At what angle should the pilot turn to plane to return from city C to city A?
20. A ship maintains an average speed of 20 kmh going from port A to port B which are
500 km apart. To avoid a storm the ship’s captain heads out a direction 15 o off course
for 8 hours.
NEED PICTURE
PORT B
500 km
Angle for course change
PORT A
a. Through what angle should the captain turn the ship to get to port B?
b. How long will it take the ship to reach port B after the captain makes the turn?
Chapter 7 Trigonometric Applications
21. A radio tower is 450 feet tall. The ground where the tower is located is on a hillside that
slopes upward at an angle of 8o. How long should two guide wires be that are attaches to
the top of the tower and secured at two points 120 feet directly above and below the base
of the tower?
NEED PICTURE
TOP OF TOWER
120 feet
120 feet
450 feet
8o
22. A hill slopes at an angle of 9o with the horizontal. From the base of the hill the angle of
elevation to the top of the 460 foot tower is 33 o. How much cable is required to connect the top
of the tower to a point at the bottom of the hill.
NEED A PICTURE
TOP OF TOWER
460 feet
24o
9o
(23−36) Choose the Law or Sines or Law of Cosines to find all the missing values of the triangle.
B
c
A
a
b
C
23.  A= 53o;  B= 62o; a = 10 inches
24.  B= 72o;  C= 29o; b = 8 cm
25. a = 8 inches; b = 10 inches; c = 15 inches
26. a = 5 m; b = 3 m; c = 7 m
27.  A= 103o;  B= 37o; c = 10 cm
28.  B= 57o;  C= 113o; a = 3 feet
29. a = 10.3 m; b = 12.2 m;  C= 83.2o
30. b = 5.9 cm; c = 13.5 cm;  A= 28.9o
31. a = 3.8 cm; b = 5.5 cm;  A= 29o30
32. a = 5.9 m; c = 3.8 m;  C=33o20
33. b = 5.9 inches; c = 3.8 inches;  C=33o10
34. a = 9.3 m; b = 7.2 m;  B= 29o40
Chapter 7.1
35. a = 8.9 feet; b = 5.2 feet;  B= 83o50
36. c = 3.8 m; a = 9.3 m;  C=79o40
37. a = 8 inches; b = 9 inches; c = 23 inches
Explain why a triangle with these measurements is not possible.
38. a = 8 feet; b = 7 feet;  A= 56o;  B= 63o
Explain why a triangle with these measurements is not possible.
Chapter 7 Trigonometric Applications
Section 7.4 Complex Numbers in Trigonometric Form
Objectives

Understanding the trigonometric form for complex numbers

Understanding multiplication and division with the trigonometric form

Understanding DeMoivre’s Theorem
To start with let’s review real quickly some of the basics about complex numbers that we talked
about in section 2.1.
A complex number is any number of the form (z = a + bi), where a and b are real numbers
and i equals
1 .
To graph a complex number you use the value for a as the x- value of a point and the value for b as
the y-value of a point. The x-axis is the real number line and the y-axis is the imaginary line. This
creates what is called the complex plane.
Imaginary axis bi
Real axis a
the point
z = 3+2i
The absolute value of a complex number is defined as follows.
To find the absolute value of a complex number (z = a + bi) you find the distance from the
origin to the point in the complex plane.
a  bi 
 a  0 2   b  0 2
 a 2  b2
For the complex number graphed above the absolute value would be 3  2i  32  22  13 .
Example 1: Absolute value
What are the absolute values of the following complex numbers?
a) −1 + 5i
b) 4 − 3i
c) −12 − 5i
Solution:
a) 1  5i
 12  52
b) 4  3i
42   3  16  9  25  5
c) 12  5i
 122   52
 26
2
THE TRIGONOMETRIC FORM OF COMPLEX NUMBERS
 144  25  169  13
Chapter 7.1
Let’s consider the general form of a complex number and draw a representative graph of it in the
complex plane. We will also identify a few other parts of the picture.
z = a + bi
r
b
θ
a
Question 1: What could the r and θ represent in this graph?
r is the distance from the origin to the complex number in the complex plane. Thus r is the absolute
value of the complex number a + bi. θ is an angle in standard position formed by the terminating side
that passes through the complex number. We can now make the following observations.
cos θ =
a
which means that
r
a = rcos θ
sin θ =
b
which means that
r
b = rsin θ
with a + bi substitute to replace a and b
rcos θ + rsin θi
r(cos θ + isin θ)
We have now found what is called the trigonometric form of a complex number.
Trigonometric form of a complex number − If you have a complex number z = a + bi
and r = a  bi and θ is an angle in standard position whose terminating side passes through
a + bi = r(cos θ + isin θ)
the complex number then the following is true.
A shortcut form for this notation is (rcis θ).
Discussion 1: Finding the Trigonometric Form
Let’s look at how we find the trigonometric form of the following complex numbers.
a)
3 i
b) 3 − 4i
c) −2 + 5i
d) 7 + 0i
a) First let’s draw a graph of the complex
z=
number in the complex plane.
r
θ
3 i
1
3
Next, we need to calculate r.
r = a  bi =
 3
2
 12  4  2
Chapter 7 Trigonometric Applications
Now, we can find θ.
The trigonometric form is
sin θ =
1

which means that θ =
3
2



3  i = 2  cos  i sin 
3
3

b) First let’s draw a graph of the complex
3
θ
number in the complex plane.
−4
r
z = 3 − 4i
Next, we need to calculate r.
Now, we can find θ.
r = a  bi =
sin θ =
 32   4 2
 25  5
4
which means that θ = −0.927
5
rad or −53.13°
The trigonometric form is
3 − 4i = 5  cos  0.927   i sin  0.927  
c) First let’s draw a graph of the complex
z = −2 + 5i
number in the complex plane.
r
5
θ
−2
Next, we need to calculate r.
r = a  bi =
Now, we can find θ.
sin θ =
5
29
 2 2  52
 29
which means that θ = 68.2°,
but that’s the reference angle in the second
quadrant so we want θ = 111.8°
The trigonometric form is
d) First let’s draw a graph of the complex
−2 + 5i = 29  cos 111.8  i sin 111.8 
r=7
z = 7 + 0i
number in the complex plane.
Now, we can find θ.
sin θ = 0 which means that θ = 0
The trigonometric form is
7 + 0i = 7  cos  0   i sin  0  
Question 2: What is the complex number in standard form given the trigonometric form

 5
2  cos 
 4


 5
  i sin  4



 ?

MULTIPLICATION AND DIVISION OF THE TRIGONOMETRIC FORM
Discussion 2: Multiplying
Chapter 7.1
Let’s look at how we multiply two trigonometric form, complex numbers.
a) 2(cos 40° + isin 40°)  5(cos 20° + isin 20°)
b) 3(cos 120° + isin 120°)  7(cos 60° + isin 60°)
a) On part (a) let’s try FOILing and collecting like terms. After we FOIL, we should
separate the i-terms from the others. Then we will see if we can apply the sum and
difference identities and then find a pattern.
Differences cos 40° cos 20° − sin 40° sin 20° = cos (40° + 20°) = cos 60°
Sums
cos 40° sin 20° + sin 40° cos 20° = sin (40° + 20°) = sin 60°.
2(cos 40° + isin 40°)  5(cos 20° + isin 20°) =
10 (cos 40° cos 20° + icos 40° sin 20° + isin 40° cos 20° + i2sin 40° sin 20°) =
10(cos 40° cos 20° + i2sin 40° sin 20°) + 10(icos 40° sin 20° + isin 40° cos 20°) =
10(cos 60° + isin 60°) which looks like 2(5)(cos (40° + 20°) + isin(40° + 20°))
b) Let’s do the same on part (b) that we did on part (a).
Differences cos 120° cos 60° − sin 120° sin 60° = cos (120° + 60°) = cos 180°
Sums
cos 120° sin 60° + sin 120° cos 60° = sin (120° + 60°) = sin 180°.
3(cos 120° + isin 120°)  7(cos 60° + isin 60°) =
21 (cos 120° cos 60° + icos 120° sin 60° + isin 120° cos 60° + i2sin 120° sin 60°) =
21 (cos 120° cos 60° + i2sin 120° sin 60°) + 21(icos 120° sin 60° + isin 120° cos 60°) =
21(cos 1800° + isin 1800°) which looks like 3(7)(cos (120° + 60°) + isin(120° + 60°))
Here is what the formula is for the product of two complex numbers in trigonometric form.
Product − If you have the product of two complex numbers
r1 (cos1  i sin 1 ) r2 (cos 2  i sin  2 ) , then the answer will be
r1r2 cos(1   2 )  i sin(1   2 )  .
[in shortcut notation
r1r2cis(1  2 ) ]
Discussion 3: Division of the Trigonometric Form
Let’s look at the division of two complex numbers in trigonometric form.
15  cos300  i sin 300 
3  cos60  i sin 60 
Let’s do an old trick and multiply the numerator and the denominator by a conjugate.
Then we will FOIL and collecting like terms. After we FOIL, we should separate the iterms from the others. Then we will see if we can apply the sum and difference identities
and then find a pattern.
Sums
cos 300° cos 60° + sin 300° sin 60° = cos (300° − 60°) = cos 240°
Differences cos 300° sin 60° − sin 300° cos 60° = sin (300° − 60°) = sin 240°.
 15  cos300  i sin300   cos60  i sin 60 
=


 3 cos60  i sin 60   cos60  i sin 60 


Answer Q1:
r would be the
distance from
the origin to
the point. θ
would be the
angle from the
positive x-axis
to the line
between the
origin and the
point.
Chapter 7 Trigonometric Applications

15 cos300 cos 60  i cos300 sin 60  i sin 300 cos 60  i 2 sin 300 sin 60

3 cos 60  i sin 60
2
2
2

=
15 cos300 cos 60  i 2 sin 300 sin 60  i  sin 300 cos 60  cos300 sin 60  

=
3 cos 2 60  sin 2 60


15 cos300 cos60  sin 300 sin 60  i sin 300 cos60  cos300 sin 60 
31
=
5 cos  300  60  i sin  300  60  = 5 cos  240  i sin  240  5cis240
Here is what the formula is for the quotient of two complex numbers in trigonometric form.
r (cos1  i sin 1)
Quotient − If you have the quotient of two complex numbers  1
, then the
r2 (cos2  i sin 2 )
answer will be
r1
cos(1  2 )  i sin(1  2 ) .
r2 
[in shortcut notation
r1
cis(1   2 ) ]
r2
Example 3: Multiplying and Dividing
What are the simplified versions of the following?
a) 4  cos72  i sin 72  3 cos 48  i sin 48
b)
4  cos72  i sin 72 
3 cos 48  i sin 48 
Solution:
12 cos  72  48  i sin  72  48  =
a) Simply add the angles and multiply the
coefficients.
12 cos120  i sin120
b) Simply subtract the angles and divide the
4
cos  72  48   i sin  72  48   =
3
coefficients.
4
cos  24   i sin  24  
3
We have multiplied and divided so let’s look at raising to a power and taking a root.
RAISING TO A POWER AND TAKING THE ROOT OF A COMPLEX NUMBER
Discussion 4: Raising to a Power
Let’s simplify the following expressions.
Answer Q2:
 2
 2
2
i

 2
2 

= 2  i 2
2
= −1 + i
2
a) r  cos  i sin  
b) r  cos  i sin  
2
a) We begin by using the
multiply property.
r  cos  i sin   r  cos  i sin   =


r 2 cos      i sin     =

r 2 cos  2   i sin  2 

3
Chapter 7.1
r  cos  i sin   r  cos  i sin   r  cos   i sin   =
b) We use the multiply
property twice.


r 2 cos      i sin     r  cos  i sin   =


r 3 cos  2     i sin  2    =

r 3 cos  3   i sin  3 

Question 3: What do you think is the simplified form of r  cos  i sin   ?
4
DeMoivre’s Theorem − For any complex number r  cos   i sin   and natural number n,
the following is true.
 r  cos  i sin    r n cos  n   i sin  n 
n
You are probably beginning to see why one might want to write complex numbers in this new form.
Example 4: DeMoivre’s Theorem
What is the value of the following?
10


 

2  cos  i sin  
4
4 
 
1  i 10  
Solution:
We apply DeMoivre’s Theorem
10
 

 
 2  cos  i sin  
4
4 
 

10 
 2
 
  
 cos 10   i sin 10   
 4
 4 


 5
25  cos 
 2


 5
  i sin  2



 

32  0  i1 
32i
Here is another fact when working with complex numbers in trigonometric form.
Complex Root Theorem − For any complex number r  cos   i sin   r  0 and natural
1
1
 
360 
360  

number n, the following is true.  r  cos  i sin   n  r n cos   k
  i sin   k


where k = 0, 1, 2, …, n − 1
Example 5: Finding the cube root of a complex number
What is the cube root of z = 1 + 0i?
Solution:
n
n 
n
n 
Chapter 7 Trigonometric Applications
First we need the complex
z = 1 + 0i = 1(cos 0 + isin 0)
number’s trigonometric form.
Now, we need to calculate the
cube root.
For k = 0
1
1
 0
360 
360  
0
1 cos0  i sin 0  3  13 cos   k
  i sin   k

3
3
3
3 


 
1
1
 0
360 
360  
0
1 cos0  i sin 0  3  13 cos   0
  i sin   0

3
3
3
3 




 1cos  0  i sin  0
=1
For k = 1
1
1
 0
360 
360  
0
1 cos0  i sin 0  3  13 cos   1
  i sin   1

3
3
3
3 


 
 1cos 120  i sin 120 
= 1 cos60  i sin 60
 1
 2
= 1  i
=
For k = 2
(ref. angel 60°)
3

2 
1
3

i
2
2
1
1
 0
360 
360  
0
1 cos0  i sin 0  3  13 cos   2
  i sin   2

3
3
3
3 




 1cos  240  i sin  240
= 1 cos60  i sin 60
 1
 2
= 1  i
=
Therefore the cube root of the
complex number 1 is
Section Summary:
3
3

2 
1
3

i
2
2
 1
3   1
3 
1  0i  1,  
i ,  
i
 2


2   2
2 

(ref. angel 60°)
Chapter 7.1
SECTION 7.4 PRACTICE SET
(1−12) Graph
1. 2 + 3i
2. 5 + 2i
3. 3 − 2i
4. 2 − 5i
5. −2 + 3i
6. −3 + 2i
7. −5 − 4i
8. −4 + 5i
9. 2i
10. −3i
11. −2
12. 3
Answer Q3:
r 4 cos(4 )  i sin(4 ) 
(13−24) Write each complex number in rectangular form
13. 3cis(60o)
14. 2(cos30o + isin30o)
15. 8(cos45o+isin45o)
16. 10cis(45o)
17. 2cis(150o)
18. 3(cos120o + isin120o)
19. 3(cos215o + isin215o)
20. 5cis(215o)
21. 4cis(210o)
22. 9(cos240o + isin240o)
23. 12(cos300o + isin300o)
24. cis(330o)
(25−36) Write each complex number in trigonometric form r(cosθ + isinθ)
(θ is in the interval [0, 360o) )
25. 5 + 5i
26. 4 + 4 i 3
27. −3 + 3i 3
28. −5 + 5i
29. −2 − 2i
30. −5 − 5i 3
31. 7 − 7i 3
32. 8 − 8i
33. −3
34. 5
35. 3i
36. −2i
37. What do you do with the arguments when you multiply 2 complex numbers in trigonometric
form?
38. What do you do with the absolute values when you multiply 2 complex numbers in
trigonometric form?
(39−46) Find the product of each pair of complex numbers.
(θ is in the interval [0, 360o) )

  

   3cis 180 
39.  2 cos30  i sin 30  3 cos90  i sin 90 

41.
40. 5cis 45

3cis 183   2cis 302 

  


43.  2 cos35  i sin 35  4 cos108  i sin108 

45.
 

42.  2 cos 210  i sin 210  5 cos70  i sin 70 




44.
9cis 305  5cis  23 
5cis 180   4cis  210 

 

46.  4 cos 270  i sin 270  3 cos 215  i sin 215 



37. What do you do with the arguments when you divide 2 complex numbers in trigonometric
form?
Chapter 7 Trigonometric Applications
38. What do you do with the absolute values when you divide 2 complex numbers in
trigonometric form?
(49−56) Find the quotient of each pair of complex numbers.
(θ is in the interval [0, 360o) )


2  cos120  i sin120 
4 cos150  i sin150
49.
 
2cis  30 
52.


6  cos93  i sin 93 
24 cos318  i sin 318
53.

 i sin150 
 
5cis 128 
30cis 217
54.
 
5cis  212 


10  cos323  i sin 323 
40 cos113  i sin113
25cis 38
55.

3  cos150
15 cos300  i sin 300
18cis 150
51.
 
3cis  45 
9cis 225
50.
56.
(57−72) Find each power and write the answer in rectangular form (a + bi)
57.

3i

4
58.  3  3i 
59.  5  5i 
5
61.
 3  3i 2 
63.
 2  2i 3 
65.
3cis(120 )
3
4
3


67.  2 cos135  i sin135 

69.
3cis(100 )
3
3
5
60.
3  3i 3 
62.
 1  i 2 
64.
 2
66.
 2cis(135 )
3
4
2  2i 2
 

5
4

68. 3 cos 45  i sin 45 
70.
 2cis(50 )
4
4
(71−80) Find all the solutions of each equation. (Leave the answers in trigonometric
form)
71. x3  1  0
72. x3  1  0
73. x3  i  0
74. x3  i  0
75. x3  8  0
76. x3  27  0
Chapter 7.1
77. x 4  1  0
78. x4  16  0
79. x 4  i  0
80. x5  i  0
81. Solve the equation x3 − 1 =0 algebraically by factoring with the difference of cubes and setting
each factor equal to zero. Compare the answer with number (71).
82. Solve the equation x3 + 1 = 0 algebraically by factoring with the sum of cubes and setting each
factor equal to zero. Compare the answer with number (72).
Chapter 7 Trigonometric Applications
Section 7.5 Polar Coordinates
Objectives

Understanding polar coordinates

Converting from rectangular to polar and vice versa

Graphing polar functions
Up to this point in the book we have graphed all functions with the Cartesian coordinate system or
what we could also refer to as the rectangular coordinate system. That is, where one moves
horizontally (x) and vertically (y) to plot points in space. But now, we are going to introduce a new
system called the polar coordinate system. In the polar system we will use distance and angles to plot
points in space.
POLAR COORDINATE SYSTEM
With the polar coordinate system the old origin is called the pole and the positive half of the old xaxis is called the polar axis. A point can be plotted with angles measured from the polar axis
(standard position angles) and with a known distance from the pole to the point.
How to plot points on a polar graph − The form of a point is (r, θ).
1) Locate the direction towards the point by using the angle (θ) in standard position.
2) Move in that direction a distance r out from the pole. (if r is positive move in the
direction of θ and if r is negative move in the opposite direction of θ)
A good way to think about this new way of plotting points is to think about a circle that has radius r
and the angle θ tells you which part of the circle to look towards from the pole. The r and θ have a
domain of  ,   .
Discussion 1: Plotting Points in the Polar Coordinate System
Plot the following polar points on the polar coordinate system.
a) (3, 30°)
4 

2
120° =
3
5
150° =
6
210° =
90° =

2
60° =
A
240° =
 4 
 5,

3 

4
3

3

6
(3, 30°)
1
7
6
d) (4, 270°)
30° =

7 
 2,

4 

180° = 
7 

c)  2,
4 

b)  5,
3 

2
3
4
5
(4, 270°)
3
270° =
2
0°
6
330° =
300° =
5
3
11
6
Chapter 7.1
Question 1: In discussion 1, in the figure there is a point labeled A. What are a couple of ways in
which we could describe this point in polar form?
CONVERTING FROM ONE SYSTEM TO ANOTHER
One of the things we will need to be able to do is to convert points from rectangular form to polar
form and vice versa. To figure out how we can do this, let’s think back to the definition of the
trigonometric functions by the coordinate definition (section 5.3).
As you look at the figure, notice all of the
connections we know between all of the
r 2  x2  y 2
P(x, y)
r
variables.
y
Pythagorean Theorem
x
 x  r cos
r
y
sin θ =
 y  r sin
r
y
tan θ =
x
cos θ =
θ
Definition
x
Definition
Definition
Notice that the rectangular form of a point (x, y) is connected to the polar form of a point (r, θ)
through trigonometric functions.
Example 1: Converting From One Form To Another
With each point in rectangular form convert them to polar form, and for each point in polar form
convert them to rectangular form.


a)  3, 
6

b) (5, −60°)
c) (−4, 4)
d)
 2, 2 3 
Solution:
a) We know that r = −3 and that  

, so we
6
simply plug those values into the equations
that connect r and θ with x and y.
x  r cos
y  r sin 
 
x  3cos  
6
 
y  3sin  
6
 3  3 3
x  3

 2 
2


 1  3
y  3   
2 2
 3 3 3 
, 

2 
 2
Thus, the point in rectangular form is
b) We know that r = 5 and that   60 , so
we simply plug those values into the
equations that connect r and θ with x and y.
x  r cos
y  r sin 
x  5cos  60 
y  5sin  60 
1 5
x  5  
2 2
  3  5 3
y  5

 2 
2


Chapter 7 Trigonometric Applications
 5 5 3 
 ,

2 
2
Thus, the point in rectangular form is
c) There are several ways to find r and θ given
r 2  x2  y 2
y
x
tan  
x and y. Let’s use the definition of tangent
r 2   4    4 
2
4
tan  
 1
4
and the Pythagorean Theorem.
r 2  16  16  32
[Since (−4, 4) is in
r  32  4 2
the 2nd quadrant]

2
3
4
3 

 4 2, 4 


Thus, the point in polar form is
d) Let’s use the definition of tangent and the
r 2  x2  y 2
y
x
tan  
Pythagorean Theorem.
2

 
r 2   2  2 3
2 3
3
tan  

2
1
2
r 2  4  12  16

[Since 2, 2 3 is in
r 4
st
the 1 quadrant]


3
 
 4, 3 


Thus, the point in polar form is
Not only can we convert points from rectangular form to polar form and vice versa, but we can
convert equations too.
Discussion 2: Converting Equations
Convert the following equations from polar form to rectangular and vice versa.
a) x 2  y 2  5
b) y  2 x 2  3
c) y  3x  7
a) As with converting points, use what
we know about the relationships
between r, θ, x and y to convert the
equations.
The polar equation is
d) r sin   2
e) r  3
x2  y 2  5
Given (circle)
r 2  x2  y 2
Known relationship
r2  5  r  5
r 5
Curves can have nice equations in polar form.
b) As with converting points, use what
we know about the relationships
between r, θ, x and y to convert the
equations.
y  2x2  3
Given (quadratic)
x  r cos , y  r sin 
Known relationships
 r sin    2  r cos 2  3
Chapter 7.1
The polar equation is
r sin   2r 2 cos2   3
Usually we want r equal to a function of θ
r(θ), but that isn’t very easy to get in this case.
c) We will do the same procedure as we
did in parts a and b.
y  3 x  7
Given (linear)
x  r cos , y  r sin 
Known relationships
 r sin    3 r cos   7
r sin   3r cos  7
r  sin   3cos   7
The polar equation is
r
7
 sin   3cos 
Notice that lines are nicer in rectangular form.
d) Replace r, and θ, with x’s and y’s.
The polar equation is
r sin   2
Given
y  r sin 
Known relationship
y2
Notice that lines are nicer in rectangular form.
e) Replace r with x’s and y’s.
r  3
Given
r 2  x2  y 2
Known relationship
9  x2  y 2
The polar equation is
Notice that circles are nicer in polar form.
GRAPHING POLAR FUNCTIONS
As mentioned earlier polar functions are of the form r(θ), where θ is the independent variable (input)
and r is the dependent variable (output). To graph a polar function we will start with an input of 0°
and continue around in a counterclockwise manner plugging in angles until we begin to repeat the
graph.
Discussion 3: Graphing A Polar Function
Let’s see what the graph of r  1  cos will look like.
Let’s begin with a table of values.
θ
r
θ
r
0°
2
135°
0.3
30°
1.87
150°
0.134
45°
1.7
180°
0
60°
1.5
225°
0.3
90°
1
270°
1
120°
0.5
360°
2
Answer Q1:
 2 
 6, 3 


5 

 6, 3 


Chapter 7 Trigonometric Applications
We will now plot our points and connect the
dots.
1
We call a curve of this type a cardioid. Any function of the form r  a  cos or
r  a  sin  will be a cardioid.
Let’s look at how
TI 83/84
we can use our
First, push the MODE key and
graphing
then arrow down and over and
calculators to
highlight Pol.
TI 86
graph a function
of this type. You
can use either
degrees or
radians, just
Second, type in your function.
make sure that
the mode menu
is set the way
you want it.
Lastly, for a good graph push
ZoomFit [0] (this takes a while to
graph) and then ZSquare [5].
Example 2: Graphing Polar Functions
What are the graphs of the following polar functions?
a) r = 2
b) r = 3sin θ
c) r = sin (3θ)
2
Chapter 7.1
a) Since there isn’t a θ in the function, this
is similar to the function (y = b), a
horizontal line. No matter which value of
θ is used as an input the radius will
always be 2. Thus we have a circle with
a radius of 2.
b) For this one let’s try a few points and see
if we can discern a pattern.
θ
r
0°
0
45°
2.12
90°
3
135°
2.12
180°
0
225°
−2.12
270°
−3
315°
−2.12
360°
0
1
2
3
Notice that when θ is in the 3rd quadrant the
radii are negative which puts the points
back in the 1st quadrant right on top of what
was already graphed.
c) Let’s plot a few points and see how
moving the three around effects the
graph.
θ
r
0°
0
15°
0.7
30°
1
45°
0.7
60°
0
75°
−0.7
90°
−1
105°
−0.7
Notice that when 60    120 the radii are
120°
0
negative which puts the points down in the
135°
0.7
3rd and 4th quadrants. Here it is graphed on
150°
1
165°
0.7
180°
0
1
the graphing calculator.
Chapter 7 Trigonometric Applications
Question 2: What is the graph of r  1  2cos ?
Section Summary:
Chapter 7.1
SECTION 7.5 PRACTICE SET
(1−16) Graph each ordered pair on a polar coordinate system.
1. (2, 30o)
2. (2, 150o)
3. (3, 120o)
4. (3, 240o)
5. (2, 225o)
6. (2, 315o)
7. (−2, 30o)
8. (−2, 150o)
9. (−3, 120o)
10. (−3, 240o)
11. (−2, 225o)
12. (−2, 315o)
13. (3, 0o)
14. (−3, 0o)
15. (−2, 90o)
16. (2, 90o)
(17−24) Covert to rectangular coordinates (use exact values).
17. (3, 45o)
18. (−3, 45o)
19. (2, 120o)
20. (−2, 120o)
21. (−2, 210o)
23. (3, −150o)
24. (−3, −150o)
22. (2, 210o)
(25−34) Convert to polar coordinates (r> 0 and 0o < θ < 360o)
25. (2, 2)
26. (−2, 2)
27. 3 3,3

28.
3
3,3

31. (−5, 0)
29.
2
2,2

32. (5, 0)
30.
 2

2,2

33. (0, 3)
34. (0, 3)
(35−42) Use your graphing calculator to convert to rectangular coordinates.
(Round answers to 3 significant digits)
35. (2, 380o)
36. (−2, 38o)
37. (−3, 111O)
38. (3, 111o)
39. (4, 235o)
41. (−5, 319o)
42. (5, 319o)
40. (−4, 235o)
(43−46) Use your graphing calculator to convert to polar coordinates.
(Round answers to 3 significant digits)
43. (3, 7)
44. (−5, 8)
45. (−7, −2)
46. (2, −9)
(47−54) Write each equation with polar coordinates.
47. x + y = 3
48. x − y = 2
49. x2 + y2 = 25
50. x2 + y2 = 9
51. x2 + y2 = 3y
53. y = −2x
54. y = 2x
52. x2 + y2 = 5x
(55−62) Write each equation with rectangular coordinates.
55. r = 3sinθ
56. r = −2cosθ
57. r2 = 16
58. r2 = 25
59. r2 = 3cos2θ
60. r2 = 2sinθ
Chapter 7 Trigonometric Applications
61. 2rsinθ + 3rcosθ = 5
Answer Q2:
62. 2rcosθ − 3rsinθ = 3
(63−68) Graph each of the following polar equations by hand.
(Show a table of values you used to create points)
63. r = 2
64. r = 4
65. r = 2cosθ
66. r = 2sinθ
67. r = 3 + 2sinθ
68. r = 2 + 3cosθ
(69−88) Graph each of the following polar equations with the graphing calculator
and state what type of graph each represents.
69. r = 3sinθ
70. r = 2cosθ
71. r = 3cos2θ
72. r = 2sin3θ
73. r = 3sin3θ
74. r = 2cos4θ
75. r = 3 + 2sinθ
76. r = 2 + 3cosθ
77. r = 2 − 3cosθ
78. r = 3 − 2sinθ
79. r = 3 + 3sinθ
80. r = 2 + 2cosθ
81. r2 = 9cos2θ
82. r2 = 4sin2θ
83. r2 = 9sin2θ
84. r2 = 4cos2θ
85. r = 2θ
86. r = −3θ
87. r = −2θ
88. r = 3θ
Chapter 7.1
Section 7.6 Parametric Equations
Objectives

Understanding The Basics of Parametric Equations

In this section we are going to talk about another way of expressing relations. In much of this book
we have expressed relations in rectangular form, some examples are y = x2, y = ex or y = sin x (these
three relations happen to be functions). In the last section we talked about expressing relations in
polar form r = f(θ). The points on a curve in rectangular form are (x, y) and the points in polar form
are (r, θ). In this new form of expressing relations, called parametric form, a point will be (f(t), g(t)),
where f(t) = x and g(t) = y.
BASICS OF PARAMETRIC EQUATIONS
As you might have noticed we will be using the rectangular coordinate plane to graph relations
expressed in parametric form. The difference being that we will be defining each variable: x and y, in
terms of a third variable t called a parameter. What is nice about this form is that it will give a curve a
direction of motion and each point will have a time reference connected to it. In many cases the
variable t will represent time and so by defining a curve in this way not only can we tell where it is
but also when it is there and in which direction it is moving. This is extremely convenient when
trying to model moving particles. Let’s work with an example and see what it is we are talking about.
Discussion 1: Graphing a Parametric Curve
Let’s see how we would graph the parametric curve: x = t + 1, y = t2 + 2t.
Let’s begin by creating a table of values
to help us create a graph.
Let’s plot the points that we found from
our table.
t
−4
−3
−2
−1
0
1
2
x
−3
−2
−1
0
1
2
3
y
8
3
0
−1
0
3
8
Chapter 7 Trigonometric Applications
From what we see from just a few points
t = −3
it might be safe to say that this is turning
t=3
out to be a quadratic. Thus the graph
would look like this. We should also
t = −2
t=2
notice that as t increased in value we go
from the left side of the curve to the right
t = −1
side.
By performing substitution we could
t=1
t=0
x = t + 1, y = t2 + 2t
prove that this is indeed a quadratic.
Solve for t in the first equation.
t=x−1
Now substitute into the other equation.
y  ( x  1) 2  2( x  1)
y  x2  2x  1  2 x  2
y  x2  1
Let’s look at how we can get our graphing calculators to graph these equations.
First, with the TI-83/84/86 you will need
to go to the MODE window and
highlight the Par and not the Func.
Next, go to the Y = window and type in
the equations
Now use ZOOM standard. Notice how it
graphed the curve. It started at the point
(1, 0) and then went to the right and up.
This happened because the default
setting is set to start with T=0.
If we change the window to have T start
with T=4 we get this graph. With the
graphing calculator we can control when
the graph begins and ends.
Use the F1 key to get t.
Chapter 7.1
Example 1: A Parametric Curve and Trigonometry
Sketch a graph of the following parametric curve, eliminate the parameter and state the name of the
resulting curve which will be in rectangular form.
x = 2cos t, y = 2sin t.
Solution:
Let’s begin by creating a table of values to help us create a graph.
t
−π
x
y
−2
0
3
4

2

4
 2
0
−2
2
 2
 2

4
0
2
0
2
2

2
3
4
0
2
 2
2
π
−2
0
Let’s plot the points and sketch what we
think the curve might look like. Also we
need to indicate a direction.
(−2, 0)
From what we see from just a few points
it might be safe to say that this is turning
out to be a circle of radius 2.
We can more easily determine the type of
x = 2cos t, y = 2sin t
curve we have if we perform substitution
Sine and cosine are more easily eliminated if
and change this from a parametric curve
we can get them to be squared.
form to a rectangular form.
x2 = 4cos2 t, y2 = 4sin2 t
x 2  y 2  4cos 2 t  4sin 2 t

 4 sin 2 t  cos2 t

 4 1  4
This confirms it. We have a circle.
x2  y 2  4
(circles will be talked about in chapter 10)
Notice that since this is a circle, as the parameter t increases you go counterclockwise around and
around forever. Let’s do another example.
Example 2: Equation in Three Forms
Sketch and convert the following parametric equation into its rectangular form. Then change it to its
polar form.
x  4sin 2 t , y  4cos2 t
Solution:
Let’s begin by creating a table of values to help us create a graph.
t
−π
3
4

2

4
0

4

2
3
4
π
Chapter 7 Trigonometric Applications
x
y
0
4
2
2
4
0
2
2
0
4
2
2
4
0
2
2
0
4
Let’s plot the points and sketch what we
think the curve might look like. Also we
(0, 4)
need to indicate a direction.
From what we see from just a few points
it might be safe to say that this is part of a
line from the point (0, 4) to the point
(4, 0). We also notice that as t increases in
value we bounce back and forth from
these two points.
Let’s convert this parametric equation
x  y  4sin 2 t  4cos 2 t

into its rectangular form.
= 4 sin 2 t  cos2 t
This confirms it. We have a line. (In this
particular case a line segment.)
Let’s now convert this rectangular form

=4
x y4
0  x  4, 0  y  4
x  y  4,
x  r cos ,
y  r sin
r cos  r sin  4
to its polar form.
r  cos  sin    4
The restriction on θ is because the
parametric function was only in the first
r
4
 cos  sin  
0  

2
quadrant.
Discussion 2: Doomsday?
The formula for the path of the earth around the sun in its rectangular form is roughly
( x  0.017) 2
y2
0.999711

 1 . In its polar form it is r 
(these formulas are created with
1
0.999711
1  0.017 cos
the units on x, y and r of Astronomical units  1Au is the distance from the earth to the sun which is
approximately 93,000,000 miles). Since the parametric form of an equation is of a rectangular nature
(x = f(t), y = f(t)) and we know that x = r cos θ and y = r sin θ, then the parametric form of a polar
equation will be x = r(t) cos t and y = r(t) sin t. So for the equation of the earth’s path in parametric

0.999711 
 0.999711 
 cos t , y   1  0.017 cos t  sin t .
1

0.017
cos
t




form we have the following x  
Now we need to adjust the trigonometric functions in the formula so that the orbit is
completed in one year. To do this we replace t with 360t, thus completing one revolution in one year.
As t goes from 0 to 1, the angle 360t, will go from 0° to 360°. Let’s graph all three of these on our
graphing calculators and confirm that they do give us the same graph.
Chapter 7.1
Rectangular:
( x  0.017) 2
y2

1
1
0.999711


y   1  ( x  0.017)2 (0.999711)
Notice that in the MODE menu, Func
is highlighted.
Polar:
r
0.999711
1  0.017 cos
Notice that in the MODE menu, Pol is
highlighted.
Parametric:


0.999711
x
 cos(360t )
1

0.017cos(360
t
)




0.999711
y 
 sin(360t )
 1  0.017cos(360t ) 
Notice that in the MODE menu, Par is
highlighted.
Now, what if we found a comet in orbit around the sun and we had determined its path to be the
equation r 
0.398
in polar form. Would it hit the earth anytime soon? Let’s graph the earth
1  0.99cos( )
and the comet together in the two forms: rectangular and polar.
Rectangular:


1  ( x  19.8) / 400 (7.96)
y   1  ( x  0.017)2 (0.999711) Earth
y
2
Comet
Chapter 7 Trigonometric Applications
Polar:
0.999711
1  0.017 cos
0.398
r
1  0.99cos( )
r
Earth
Comet
With the rectangular and polar forms we just get the graph of the path that the earth and
comet will follow. That isn’t enough to know if they will collide only that they might collide. But,
with the parametric form, not only can we get the path that the object will follow, but we can also
incorporate into the formula the speed at which it is moving. This lets us know where it will be, and
when it will be there, for any particular point along its path. That aspect is very important in this type
of problem!
Thus in relation to the speed and location of the earth to the comet, if we know that the


(t  32)
comet’s parametric formula needs a parameter of  29
 180  , while earth needs a
 1.3 1064



parameter of 360(t + 0.1467), we get the following graphs.
Parametric:
Comet






 
0.398
(t  32)
x
 180 
 cos  29

64


 1  0.99cos  29 (t  32)  180    1.3  10
 1.3  1064












 
0.398
(t  32)
y 
 180 
 sin  29

64


 1  0.99cos  29 (t  32)  180    1.3  10

64
 1.3  10





Earth


0.999711
x
 cos(360(t  0.1467))
1

0.017cos(360(
t

0.1467))




0.999711
y 
 sin(360(t  0.1467))
1

0.017cos(360(
t

0.1467))


Chapter 7.1
In the graph we see the earth’s orbit and we can see the
comet approaching.
Let’s zoom in on that part of the graph and see if
Doomsday is approaching. The point highlighted in the
graph is the location of the earth.
It looks as though the earth will survive this time around
since the comet has passed just in front of the earth. The
point highlighted in the graph is the location of the comet.
Note: the parametric formulas used aren’t exactly the ones used by NASA but are a good
approximation and make for a good example of how parametric functions can be useful.
Section Summary:
Chapter 7 Trigonometric Applications
SECTION 7.6 PRACTICE SET
(18) Graph by plotting points the plane curve given by the parametric equation.
Then find an equivalent rectangular equation.
1
y  t,
3  t  6
1. x  2t 1,
2. x  3t  2,
y  3t  1, 2  t  5
3
3. y  t 2  1,
x  t  2, 4  t  4
1
4. y  2t 2  3, x  t ,
2
5. x  t ,
y  3t 1, 4  t  4
6. x  t  6,
y  2t  1, 2  t  3
7. x  t 2 ,
y  t  3,
8. x  t 2 ,
y
2  t  5
6  t  6
1
t  3, 4  t  12
2
(924) Graph each of the following parametric equations with the graphing calculator.
Then find an equivalent rectangular equation and graph the rectangular equation with the
graphing calculator to show they create the same graph.
9. x  t 2  3,
y  t 2  1,
4  t  4
10. x  t 2  2,
y  t 2  5,
4  t  4
11. x  t 3 1,
y  t 3  3,
4  t  4
12. x  t 3  3,
y  2t 3 1,
2  t  2
13. x  3sin t ,
y  3cos t ,
0  t  2
14. x  2sin t ,
y  2cos t ,
0  t  2
15. x  2sin t ,
y  cos t ,
0  t  2
16. x  sin t ,
y  2cos t ,
0  t  2
17. x  sin t ,
y  csc t ,
0t 
18. x  cos t ,
y  sec t ,


t 
2
2
19. x  2sin t  1,
21. x  3  sec t ,
23. x  ln t,
y  2  2cos t ,
0  t  2
y  1  2 tan t , 0  t 
y  t2, 0  t  

2
20. x  2cos t  1,
y  2  2sin t ,
22. x  1  2cot t, y  3  csc t,
24. x  et ,
y  et ,   t  
0  t  2
0 t 
Chapter 7.1
(2526) PROJECTILE MOTION
Neglecting air resistance, the following equations will describe the path of a projectile propelled
upward at an angle of  with the original horizontal height h, in feet, at an initial speed vo in feet
per second.
x   vo cos   t , y  h   vo sin   t  16t 2
For the following problems find:
a. Find the parametric equation that gives the position of the object at time t, in seconds.
b. What is the height of the object at 1 sec, 2 sec, 3 sec?
c. What is the maximum height of the object?
d. How long is the object in the air?
e. What is the horizontal distance the object travels?
f. Graph the equation with the graphing calculator and use the calculator to check the answers
for c, d, and e.
25. A baseball is hit upward from a height of 3 feet with and initial speed of 150 ft/sec at an angle of
30o with the horizontal.
26. An object is launched from the ground with an initial speed of 300 ft/sec at an angle
of 60o to the horizontal.
27. A golf ball is hit from ground level with an initial speed of 180 ft/sec at an angle of 35 o with
the horizontal.
28. A cannon ball is fired from a cannon with an with its muzzle 3 feet about the ground at an angle
of 50o to the horizontal and initial speed of 800 ft/sec.
29. If the baseball from problem 25 was his with and angle of 60 o with the horizontal, how far does
it travel? If the angle was 45o how far does the base ball travel? What do you think is the
relationship with the distance and the angle?
30. If the golf ball from problem 27 is his with and angle of 50 o with the horizontal, how far does
the golf ball travel? If the angle is 75o how far does the golf ball travel?
Chapter 7 Trigonometric Applications
Section 7.7 Vectors
Objectives

Understanding The Basics of Vectors

Performing operations on Vectors

Understanding the Unit Vector form of Vectors

Understanding the Trigonometric definition of Vectors
In this section we continue our discussion about the different areas of mathematics where
trigonometric functions are used. So far in this chapter we have discussed their use in problems
involving triangles (Law of Sine and Cosine), complex numbers, and polar coordinates. Now we will
turn our attention to the topic of Vectors.
BASICS OF VECTORS
Many situations in life can be expressed by real numbers. For example, your height, your weight, the
distance you need to travel to get to school, how much gas your car can hold, etc. But there are
situations where you need to not only know how much or how big or how strong but you also need to
know in which direction. A real number doesn’t help us in these types of situations. When we need to
be able to express not only the magnitude of something but also its direction we need what are called
vectors. A vector is a directed line segment. The length of the line segment expressed the magnitude
and the direction of the vector, well, expresses its direction. Therefore a vector has a starting point
(initial point), an ending point (terminating point), a length (magnitude), and a direction.
Terminating Point (Q)
Initial Point (P)
Vector PQ (v)
One way of labeling a vector is to use the points that create them. In the above figure the vector starts
at point P and goes to point Q, hence the labeling PQ . The arrow over the letters indicates that this
symbol is representing a vector. We also use lowercase bold letters at the end of the alphabet like v,
u, and w, to label vectors.
Question 1: What do you think will need to be true about two vectors in order for them to be
equivalent?
Chapter 7.1
Discussion 1: Equivalent Vectors
Let’s look at two vectors and decide if they are equivalent or not.
P(−2, 4)
Q(2, 1)
R(1, −1)
S(5, −4)
First let’s evaluate the magnitude (length)
Magnitude is distance or length so,
and direction of vector PQ .
length of PQ = (2  (2))2  (1  4)2 = (4)2  (3)2
= 16  9  25 = 5
Direction is slope so,
slope =
1 4
3

2  (2) 4
Next let’s evaluate the magnitude (length)
Magnitude is distance or length so,
and direction of vector RS .
length of RS = (5  1)2  (4  (1))2
= (4)2  (3)2 = 16  9  25 = 5
Direction is slope so,
slope =
4  ( 1) 3

5 1
4
Since both vectors, PQ and RS , have the same length and same direction (slope), we consider
them to be equivalent. Notice that with the slope we took the second point minus the first. If
you don’t do this your slope would suggest that the vector is pointing in the opposite direction.
If the answer to the slope of this problem was
3
that would suggest that we move up 3 and
4
then left 4 which is the opposite of what it does
3
down 3 right 4 (look at arrows in picture).
4
Here are a few terms and definitions.
Equivalent Vectors − Two vectors are equivalent if, they have the same magnitude and the
same direction.
Magnitude − If you have points P(p1, p2) and Q(q1, q2) that form vector PQ then its
magnitude is written PQ and is calculated using the formula PQ  (q1  p1)2  (q2  p2 )2 .
When working with vectors think second point minus the first point.
Chapter 7 Trigonometric Applications
Direction − For the moment we will use the slope of the vector to help us determine
direction.
Unit Vector − A vector whose magnitude is one, is called a unit vector.
Zero Vector − A vector whose magnitude is zero, is called a zero vector.
We need a convenient way of expressing vectors. It is too cumbersome to list the two endpoints. One
way to make life easier is to use what is called the component form of a vector.
Component form of a Vector − If we have two points that describe a vector P(p1, p2) and
Q(q1, q2) then, this is how we will express the vector.
PQ = q1  p1, q2  p2 = v1, v2 = v
and the magnitude v = v12  v22
The components v1 and v2 describe how far it is horizontally and vertically respectively, from the first
point to the second point. Thus we have a simply way of expressing all vectors and any two vectors
that are equivalent will now turn out to have the exact same v1 and v2 values. You see, the length and
direction of a vector is what is important and not where it begins.
Example 1: Component Form of a Vector
What is the component form of the following vectors?
a) v = PQ , P(3, −2), Q(4, 3)
b) v = PQ , P(−2, 1), Q(−7, −1)
Solution:
a) We simply need
q1  p1, q2  p2 = 4  3, 3  (2)
to substitute the
(1, 5)
= 1, 5
values of our two
points into our
formula.
b) Do the same as in
The vector moves one unit to the right
and 5 up.
q1  p1, q2  p2 =
part (a).
Answer Q1:
Same direction
and same
length
 7   (2),  1  1
= 5,  2
The vector moves five unit to the left and
2 down.
(−5, −2)
Chapter 7.1
 v2 
 you get the slope of the vector. There
 v1 
You should notice that by dividing the two components 
are other forms used to write a vector, one of which uses trigonometric functions, but for now let’s
use this component form and learn about performing basic operations on vectors.
OPERATIONS ON VECTORS
Let’s discuss scalar multiplication first along with addition. Scalar multiplication is where we
multiply a vector by a real number. The result is that the vector shrinks or grows in length but the
direction is left unchanged.
Example 2: Scalar multiplication
What would be the component forms of the following vectors?
a) v = 2 1, 3
b) v =
2
10,  5
5
c) v = −3 2,  5
Solution:
a) Scalar multiplication is quite easy; we
2 1, 3 = 2, 6
simply multiply every component of the
The vector now goes farther to the left and up
vector by the multiplier.
more, double the amount to be precise.
b) In this case when we multiply by the
fraction
2
we will be making the vector
5
shorter.
c) This time the multiplier is going to cause
2
10,  5 = 4,  2
5
Now the vector does not go as far to the right
or as far down.
−3 2,  5 = 6, 15
the original vector to change direction and
The vector now goes 6 to the left and 15 up
get longer in length.
instead of right 2 and down 5.
Question 2: How will multiplying the vector 4,  3 by the number −1 change the original vector?
You should have noticed that when we multiply by a value larger than one or smaller than negative
one the magnitude of the vector is increased. On the other hand if we multiply by a value between
negative one and positive one the magnitude of the vector is decreased.
Now, adding two vectors would be like hiking. If you took a hike that required you to use
two different trails you would start by going down the first trail until you reached its end. Then you’d
begin going down the second trail until you reached its end and then you’d be done with your hike.
This is how vectors work. If we look at how we add two vectors graphically we would do just like
our hike. We’d graph the first vector and then start drawing the second vector where the first one
ended. After that we draw a vector from the start of the first one to the end of the second one and that
Chapter 7 Trigonometric Applications
would be the answer (vector) of the two vectors added together. Let’s illustrates this with an
example.
Discussion 2: Graphical addition of two vectors
Let’s graph the two vectors v1 = 2, 7 and v2 = 5,  1 , and see what the sum of the two would look
like graphically (v1 + v2 = v).
We begin by drawing the first vector (v1). We starting at the
v2
origin and then going to the point that is 2 units left and 7 up.
Next draw the second vector (v2) by starting at the end of the
v1
first vector (point (−2, 7)) and then go to the point that is 5 units
v
to the right and 1 down.
Lastly, we draw a vector from the origin to the end of the
second vector and that will be the sum of the two. This vector v
might be thought of as the shortcut for getting from the start of
the hike to the end of the hike in a direct path and not by
following path v1 and then path v2.
Algebraically, we simply add the first components together and then the second components together
and the resulting values describe the new vector created by the addition of the two.
(2)  5, 7  (1)  3, 6
v1 + v2 = v
Question 3: What do you think is the process used algebraically to subtracting two vectors?
In answering this last question hopefully you thought back to how you were taught to subtract to
whole numbers. With whole numbers you should have been taught to change the sign on the second
number and then add instead of subtract (5 − 8 = 5 + (−8) = −3).
Let’s subtract the two vectors that we just added in the last example.
Example 3: Subtracting Vectors
What is the difference between the two vectors v1 = 2, 7 and v2 = 5,  1 (v1 − v2 = v)?
Solution:
v1 − v2 = v1 + (−v2) = v
(2)  5, 7  (1)  (2)  (5), 7  1  7, 8
Let’s see what this looks like graphically.
v1 − v2 = v1 + (−v2) = v
v1 + (−v2)
v1 − v 2
Chapter 7.1
−v2
v1
v1
v
v
v2
Notice that both red vectors have the same magnitude (length) and the same direction.
Graphically v1 + (−v2) follows exactly with what we did graphically when we added two vectors. We
take v1 and add to it (−v2), the second vector but going in the opposite direction. Graphically v1 − v2 is
not so intuitive but as with finding the component form of a vector given two points you drawn v
such that it goes from the end of the second vector to the end of the first. Generally, when you
subtract to determine direction we want to start with the second item and then use the first. When
finding components we take the second point minus the first likewise when we want to graph v1 − v2
we draw from the second vector to the first.
Question 4: What is the difference between v1 = 3,  4 and v2 = 1,  2 (v1 − v2 = v)?
Properties of Vectors
Since vector operations are dealing with real numbers, all of the usual properties will hold true.
Given vectors u, v, w, 0 (zero vector) and constants c, d, the following are true.
Commutative
u+v=v+u
Distributive
c(u + v) = cv + cu
Associative
u + (v + w) = (u + v) + w
Distributive
(c + d)u = cu + du
Identity
u+0=u
Associative and Commutative
Inverse
u + (−u) = 0
Magnitude of a vector
cu = c u
(cd)u = c(du) = d(cu)
Scalar Identity
1u = u
Zero scalar
0u = 0
Zero vector
c0 = 0
UNDERSTANDING THE UNIT VECTOR FORM OF A VECTOR
Now we need to move on to another way of defining vectors. But first we need to define a couple of
terms.
Unit Vector − A unit vector is one that has a magnitude (length) of one.
Horizontal Unit Vector − The vector 1, 0 labeled i (thus i = 1, 0 ). i  1
Vertical Unit Vector − The vector 0, 1 labeled j (thus j = 0, 1 ). j  1
Answer Q2:
Multiplying by
−1 will make
the vector go
in the opposite
direction and
be 4, 3 .
Chapter 7 Trigonometric Applications
This will give us a way of defining a vector in terms of two unit vectors. This is similar to our
component form of a vector which gave us how far to more horizontally and vertically to get from
the starting point to the ending point of a vector. Everything should work similarly to what we’ve
already done.
The Unit Vector Form of a Vector − The unit vector form of a vector v = a , b is:
v = ai + bj , where i and j are the horizontal and vertical unit vectors.
Example 4: Working with the Unit Vector form of a Vector
What are the results of the following questions?
a) Draw the vectors i, j and v = 3i − 4j.
b) Convert vector u =  2, 5 to its unit vector form and sketch a picture.
c) Find −2u − 3v given u = −4i + 5j and v = i − 7j.
a) To draw the vector i simply draw a vector
from the origin to the point (1, 0). To draw
j
the vector j simply draw a vector from the
i
origin to the point (0, 1). To draw the
u = 3i − 4j
vector v = 3i − 4j draw a vector from the
Answer Q3:
Do just like
with addition
but use
subtraction.
First
component
minus the first
component of
the second
vector and
then second
component
minus second
component of
the second
vector.
origin to the point (3, −4)
b) The component form tells us how far we
u = −2i + 5j
move horizontally and vertically so we just
need to place an i and a j after the
appropriate numbers and add the two new
vectors together.
5j
u = −2i + 5j
−2i
c) We simply combine the like vector parts
together to find the resulting new vector.
−2u − 3v = −2(−4i + 5j) − 3(i − 7j)
= −8i + 10j − 3i + 21j
= (−8i − 10i) + (10j + 21j)
= −18i + 31j
TRIGONOMETRIC DEFINITION OF VECTORS
Now to answer that question you’ve been thinking about this whole section; “What does
trigonometry have to do with vectors?” Well, we can, yet again, define vectors in another way. This
third way of defining vectors is popular with the science community, particularly Physics.
Discussion 3: Trigonometric form of a vector
Chapter 7.1
If we put a vector in standard position then we can see the angle formed between it and the positive
x-axis. Therefore if we know that a vector has a magnitude of 5 and that it forms an angle of

with
6
the x-axis we will be able to define the vector in a useful way. Let’s see why this works.
We can see that our vector with its horizontal
and vertical components form a right triangle.
Thus we know, from trigonometry, that
sin
 b
 a
= and that cos = . Therefore, the
6 5
6 5
vertical component is b = 5sin
5

and the
6
horizontal component is a = 5cos
a

. So, we
6

6
b
5 3 5
 
 
v = 5cos   , 5sin   =
,
2 2
6
6
get the following answers for our vector in its
component form and unit vector form.
v = ai + bj = 5cos
or


5
5 3
i + 5sin j =
i+ j
6
6
2
2
From this example we will now generalize the trigonometric form of a vector.
Trigonometric form of a Vector − Given a vector’s magnitude v and direction, in terms of an
angel between the vector and the positive x-axis, we can define the vector (v) in two ways:
v = v cos θ i + v sin θ j or
2,  2
v = v cos , v sin 
where the horizontal component is v cos θ and the vertical component is v sin θ.
Discussion 4: Conversion of Vector Forms
Let’s see how to convert from the component or unit vector form of a vector to the trigonometric
form given the vector v =
3, 1 .
To convert to the component form just
substitute.
To convert to the trigonometric form you
will need to figure out θ. To find θ you need
to realize that v = 3 i − j tells us we go
horizontally 3 units and vertically −1 units.
Thus by dividing these two numbers we will
have found the slope of the vector.
v = a , b  v = ai + bj
v=
3i−j
Slope of vector =
1
3
Answer Q4:
3  1, 4  (2)
vertical change
horizontal change
Chapter 7 Trigonometric Applications
Let’s do an example from Physics.
Example 5: A Ramp
What is the force pulling the crate down the ramp and into the ramp if the ramp has an incline of 20°
and the weight of the crate is 250 pounds?
Solution:
Let’s drawn a picture and label the important
parts.
Ramp
Crate
Force down
the ramp (d)
θ
20°
Ground
Force due to
gravity (g)
Force into
the ramp (r)
From what we know from geometry, we see that θ is the complement of the 20° angle (if we
move the arrows down notice how the force due to gravity and the force down the ramp would
form a triangle that would be similar to the triangle formed by the ramp). Thus θ = 70°.
Let’s drawn a picture of just our force vectors
r
in a slightly different orientation. We know
g
that the force due to gravity in this problem is
the weight of the crate which is 250 pounds.
70°
Thus we know its magnitude and angle, so we
d
can find the component vector’s magnitudes.
The magnitude of d would be the horizontal
d = 250 cos(70°)  85.5 lbs
component of g and r would be the vertical
r = 250 sin(70°)  234.9 lbs
component.
Therefore, we have found that there is a force of 85.5 lbs trying to pull the crate down the
incline and a force of 234.9 lbs pulling the crate into the ramp.
Example 6: Wind and Airplanes
If town B is north of town A and there is a 40 mph wind blowing east, in what direction will an
airplane need to head in order to go from town A to town B if the airplane travels at the rate of 550
mph?
Solution:
Let’s drawn a picture and
w = Wind speed = 40 mph
label the important parts.
Town B
a = Airplane speed = 550 mph
v = Resulting Vector speed
θ
Town A
East
Chapter 7.1
We know that the airplane will need to head a little to the west in order to compensate for the
wind blowing to the east. We will let v be the vector from town A to town B. We know that we
want v = a + w (the vectors of, the airplane plus the wind).
We know that the vector for the wind
w = w cos(0) i + w sin(0) j = 40i
would be
(the angle between east and the wind is 0°)
We know that the vector for the airplane
a = a cos( ) i + a sin( ) j = 550cosθ i + 550sinθ j
would be
(the angle between east and the wind is θ)
We know that the vector between the two
v = a + w = (550cosθ i + 550sinθ j) + 40i
towns would be
= (550cosθ + 40)i + 550sinθ j
But, we also know that town B is directly
Hence,
north of town A. Therefore, the vector (v)
(550cos θ + 40)i = 0i
must to equal to cj, there is no component
550cos θ + 40 = 0
in the horizontal direction.
550cos θ = −40
cos θ =
40 4

550 55
 4 
θ = cos 1    94.17
 55 
It appears that our airplane needs to head in a direction which is 94.17° from east. Or in terms
of bearings, N 4.17° W.
Section Summary:
Chapter 7 Trigonometric Applications
SECTION 7.7 PRACTICE SET
(112) For a vector between the two points A and B:
a.
b.
c.
Write the component form of the vector between A and B. (ai + bj)
Find the magnitude of the vector between A and B.
Write the trigonometric form of the vector between A and B. ( v cos i + v sin j)
1. A = (3, 5)
B = (7, 8)
2. A = (2, 3)
4. A = (3, 2)
B = (1, 5)
5. A = (2, 3)
7. A = (5, 2)
B = (14, 10)
8. A = (2, 1)
10. A = (2, 3)
B = (5, 3)
B = (10, 8)
B = (4, 11)
B = (5, 3)
11. A = (2, 1)
B = (5, 2)
3. A = (1, 3)
B = (13, 8)
6. A = (3, 2)
B = (15, 3)
9. A = (3, 1)
B = (3, 4)
12. A = (8, 2)
B = (10, 5)
(1324) For each of the vectors find u + v, u  v, 3u  2v
13. u = 2,5
14. u = 3, 2
v = 3,9
v = 4,1
16. u = 5i  2j v = 2i  8j
17. u = 2, 5
19. u = 3i  7j v = 3i  5j
20. u = 4i  5j v = 2i + 7j
21. u = 2,5
22. u = 5, 2
23. u = 3i  2j v = 3i  2j
24. u = 2i  5j v = 2i + 5j
v = 5, 2
v = 8, 1
15. u = 3i + 5j v = 4i + 2j
18. u = 5, 7
v = 2,11
v = 2, 5
(2542) Give each vector in trigonometric form. ( v cos i + v sin  j)
25.  = 30o, v  4
26.  = 120o, v  3
27.  = 145o, v  2
28.  = 60o, v  5
29.  = 240o, v  3
30.  = 225o, v  6
31.  = 300o, v  4
32.  = 330o, v  7
33.  = 0o, v  2
34.  = 90o, v  3
35.  = 270o, v  4
36.  = 180o, v  5
37.
(3,3)
 2, 2 3 
38.

39.



3,1
40.

(−4,−4)
θ
Chapter 7.1
41.
42.
θ

θ

5
3, 1
(43−54) Write each of the following vectors in trigonometric form.
( v cos I +
43. 2 3 i + 2j
44.
46. 2 3 i + 2j
47. −2i − 2 3 j
48. 2 2 i − 2 2 j
49. 3 3 i − 3j
50. 3i − 3 3 j
51. 3i
52. 4 j
53. −2 j
54. −5i
2i+
2j
2, 5 2
45.  2 i +

v sin 
i)
2j
(55−66) Write each of the following vectors in component form. (ai + bj)
55. 3cos 30o i + 3sin 30o j
56. 2cos 60o i + 2sin 60o j
57. 5cos 120o i + 5sin 120o j
58. 4cos 150o i + 4sin 150o j
59. 2cos 215o i + 2sin 215o j
60. 3cos 210o i + 3sin 210o j
61. 5cos 300o i + 5sin 300o j
62. 4cos 315o i + 4sin 315o j
63. 3cos 90o i + 3sin 90o j
64. 2cos 270o i + 2sin 270o j
65. 4cos 180o i + 4sin 180o j
66. 5cos 0o i + 5sin 0o j
67. A pilot is flying a plane from city A to city B. The bearing of the plane from city A
to city B is N 20o W and the wind speed is 10 mph due east. The plane’s average speed
is 350 mph and the it takes the plane 3 hours to fly from city A to city B.
a. What is the distance from city A to city B?
b. What is the bearing from city A to city B?
Need Picutre
wind B N
plane
W
E
A
S
Chapter 7 Trigonometric Applications
68. City A is 520 miles from city B and the bearing from city A to city B is N 32o E. The wind
speed is 8 mph.
a. What bearing must the plane fly to fly from city A to city B in 2 hours?
b. How many miles has the plane flown to get from city A to city B in 2 hours?
Need picture
wind B
N
W
plane
A
E
S
69. A ship leaves port traveling on a bearing of N 28 o E from the port and travels 20 miles on
this bearing. The ship then turns due east and travels 12 miles.
a. How far is the ship from port?
b. What is the bearing from port to the ship?
Nee Picture
N
W
Ship
Port
E
S
70. A ship leaves port traveling on a bearing N 33 o E from the port and travels 25 miles.
The ship then turns due west and travels 42 miles.
a. How far is the ship from the port?
b. What is the bearing from the port to the ship?
Need Picutre
N
Ship
W
Port
S
E
Chapter 7.1
71. A automobile is parked on a street in San Francisco and the street has an incline
of 14o. The automobile weighs 4,850 pounds, what force is required to keep the
automobile from rolling down the street into the ocean?
Need a picture
street
automobile
ocean
72. An automobile is parked on a ramp that has an incline of 8 o. The automobile weighs 4,200,
what force is required to pull the automobile up the ramp?
Need a picture
automobile
ramp
73. A plane flies at a constant rate of 600 mph with a bearing of S 45 o W. The velocity of the
jet stream is 80 mph from the west.
a. What is the actual speed of the aircraft?
b. What is the actual bearing of the aircraft?
Need picture
N
W
E
S
74. A boat is crossing a river that is ½ km wide. The boat can maintain a constant speed of
18 km per hour and the current is a constant 2 km per hour.
a. What angle should the boat travel to get to a point directly across the river/
b. How long will it take the boat to cross the river to that point?
Need picture
current
boat
Chapter 7 Trigonometric Applications
Section 7.8 Dot Product
Objectives

Understanding Dot Product

In this sec
Section Summary:
Chapter 7.1
SECTION 7.8 PRACTICE SET
(112) For each pair of vectors find:
a. u∙v
b. The angle  between u and v
1. u = 2i + j
v = 2i  j
2. u = 3i +2 j
v = 3i 2 j
3. u = 2i 5 j
v =3i + 2j
4. u = 3i  4j
v = 3i + 4j
5. u = 5i + 2j
v = 3i + 4j
6. u = 3i + 5j
v = 4i + 7j
7. u = 5i  7j
v = 4i  8j
8. u = i  4j
10. u = 3,7
v = 1,9
11. u = 2, 3
12. u = 1, 8
v = 7i  j
9. u = 2,3
v = 5, 4
15. u = 5, 4
v = 15,12
v = 5, 4
v = 3, 1
(1320) Show each pair of vectors are parallel
13. u = 5i − 2j
16. u = 3,1
v = −10i + 4j
v = 15,5
19. u = 1, 8
v = 3, 1
14. u = 2i + 3j
17. u = 2i + j
20. u = 1, 8
v = 8i + 12j
v = 2i  j
18. u = 2i + j
v = 2i  j
v = 2i  j
v = 3, 1
(2126) Show each pair of vectors are perpendicular
21. u = 2i + j
v = 2i  j
22. u = 2i + j
v = 2i  j
23. u = 2i + j
24. u = 2i + j
v = 2i  j
25. u = 2, 3
v = 3, 2
26. u = 5, 4
v = 4,5
27. Find a vector parallel to 3i − 2j. (Need just one out of an infinite number)
28. Show ai + bj and kai + kbj, where k is any real number, are always parallel.
(Challenge problem)
29. Find a vector perpendicular to 2i − 5j. (Need just one out of an infinite number)
29. Show ai + bj and −bi + aj are always perpendicular.
(31−38) Find work , w, performed when the given force, F, is applied to an object and the resulting
motion is represented by the displacement vector d. Assume the force is in kilograms and
the displacement is measured in meters.
31. F = 20i + 10j
d = 25i + 5j
32. F = 25i − 12j
33. F = −15i + 10j
d = 25i + 5j
34. F = −30i + 19j
35. F = 53i − 38j
d = −52i − 27j
36. F = 78i − 59j
37. F = 85i
d = 8i
38. F = 41j
d = 12i + 9j
d = −25i − 5j
d = 49i + 9j
d = 83i
Chapter 7 Trigonometric Applications
39. A postal worker pushes a package across the floor of the work room in the post office. The
worker moves the package 50 feet across the floor by exerting a force of 45 pounds downward
at an angle of 25 degrees with the horizontal. How much work has the postal worker done?
Need a Picture
40. A librarian pushes a box of books across the floor of the library. The librarian moves the box
35 feet across the library floor by exerting a force of 35 pounds downward at an angle of
23 degrees with the horizontal. How much work has the librarian done?
Need a Picture
41. An automobile has run out of gas and the driver pushes the automobile down a level street 500
feet. If the driver pushes at an angle of 18 degrees with the horizon, how much work has the
driver done?
Need a Picture
42. A dog pulls a sled 1,000 feet. If the dog exerts a force of 18 pound and the leash attached
to the dog forms 15 degrees with the horizontal, how much work has the dog done?
Need a Picuture
Chapter 7.1
Chapter 7 Review
Topic
Section
Keys




CHAPTER 7 REVIEW HOMEWORK
Section 7.1
Section 5.1 Geometry Review
CHAPTER 7 EXAM