Download Right Triangle Trigonometry Packet

Right Triangle Trigonometry
for College Algebra
B
c
A
opposite
a
=
hypotenuse c
adjacent
b
cos A =
=
hypotenuse c
opposite a
tan A =
=
adjacent b
sin A =
a
C
b
opposite
b
=
hypotenuse c
adjacent
a
cos B =
=
hypotenuse c
opposite b
tan B =
=
adjacent a
sin B =
Contents
I.
II.
III.
IV.
Background and Definitions (exercises on pages 3-4)
The Trigonometric Ratios (exercises on pages 6-7)
Applications (exercises on pages 9-10)
The Law of Sines and Cosines and More Applications (exercises on pages 14-15)
I.
Background and Definitions
The word trigonometry is derived from the Greek words “trigon,” meaning triangle and
“metry,” meaning measurement. For every triangle, we can measure the sides and the
angles. We use standard linear units of measure for the sides. The measure of an angle is
defined by the amount of rotation from its initial side to its terminal side.
Terminal Side
Vertex
Initial Side
For angles, we will use the degree for our standard unit of measure.
One degree is the measure of an angle equivalent to a rotation of
complete revolution about its vertex.
1
360
of a
We can further subdivide one degree into minutes and seconds as follows:
1 degree (1ο ) = 60 minutes (60' )
1 minute (1' ) = 60 seconds (60" )
EXAMPLE:
EXAMPLE:
Express 17.23ο in degrees, minutes, and
seconds.
Express 40 ο20'50" using degrees only.
SOLUTION:
SOLUTION:
17.23ο
40 ο20'50"
= 17 ο + 0.23ο
= 40 ο + 20' (1ο / 60' ) + 50" (1ο / 3600" )
= 17 ο + (0.23ο )(60' / 1ο )
≈ 40.34722
= 17 ο + 13.8'
= 17 ο13'+ (0.8' )(60" / 1' )
= 17 ο13'+48"
= 17 ο13'48"
2
At this point, we should remind ourselves what we already know about right triangles,
triangles with a 90 ο angle.
B
c
a
A
C
b
First, we may recall that ∠A + ∠B + ∠C = 180 ο . So if ∠B = 35 ο in the above triangle,
then ∠A = 180 ο − (90 ο + 35 ο ) = 55 ο . The second thing we recall about right triangles is
our oldest theorem, the Pythagorean Theorem. This theorem states that the sums of the
squares of the short sides of the triangle is equal to the square of the hypotenuse; i.e.,
a 2 + b 2 = c 2 . So if c = 13 and a = 5 , then b = 13 2 − 5 2 = 12 .
EXERCISES:
1. Convert the following degree measures to degrees, minutes, and seconds.
(a) 94.73ο
(b) 15.15 ο
2. Convert the following measures to degrees only.
(a) 20 ο19'18"
(b) 149 ο39'29"
3. For each triangle, find the unknown side.
(a)
(b)
5 cm
10 cm
20 in
11 in
3
4. Why is a circle divided into 360 ο ? This number appears to be arbitrary; why
not use 400 or for that matter 4? Use the internet to search out the answer. Be
cautious, as the internet is becoming famous for false information.
II. The Trigonometric Ratios
B
You may have noticed previously, that if
we know two angles, we may find the
third, and that if we know two sides, we
may find the third. What if we only know
two angles and one side? How might we
find the other sides? This is where we
begin our study of right triangle
trigonometry.
A
c
a
b
C
There are three trigonometric ratios that we will use for the basis of our study (there are
three others). These are called sine (sin), cosine (cos), and tangent (tan). In what
follows, the hypotenuse is always the longest side of the triangle, the opposite side is the
side opposite the angle, and the adjacent side is the side next to the angle. We have the
following definitions, based on the above triangle:
opposite
a
=
hypotenuse c
adjacent
b
cos A =
=
hypotenuse c
opposite a
tan A =
=
adjacent b
sin A =
opposite
b
=
hypotenuse c
adjacent
a
cos B =
=
hypotenuse c
opposite b
tan B =
=
adjacent a
sin B =
Make special note that the opposite and adjacent sides depend on the angle in question.
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EXAMPLE:
In the right triangle shown below, find sin A, cos A, tan A, sin B, cos B, and tan B .
C
4
SOLUTION:
A
sin A =
3
5
sin B = 4 5
cos A = 4 5
tan A =
cos B =
tan B = 4 3
3
5
3
4
5
3
B
In the previous example, note the relationship between sin A and cos B . This similar
relationship also occurs in the other ratios. Now you may recall that we earlier noted that
there are three more trigonometric ratios. These are based on the reciprocals of our three
main ratios. They are called secant (sec), cosecant (csc), and cotangent (cot) and are
defined as follows:
secθ =
1
hypotenuse
=
cos θ
adjacent
cscθ =
1
hypotenuse
=
sin θ
opposite
cot θ =
1
adjacent
=
tan θ opposite
We will focus our attention on sine, cosine, and tangent in this course, but you may find
these identities useful in subsequent courses.
At this point, we must discuss the role of calculators in our brief study of trigonometry.
Change your calculator to degree mode. To do this, press the mode button and highlight
degree if radian is currently highlighted. Now test this by entering sin(35) . You should
get about 0.57358. If you got –0.42818, you are still in radian mode.
You should see three buttons labeled sin, cos, and tan. We will use these buttons to find
unknown sides, given an angle and a side of a triangle. Above these buttons, you will
find sin −1 , cos −1 , and tan −1 . To use these, you will first press the 2nd or inverse button.
The key word here is inverse. These will “undo” what sin, cos, and tan “do,” much in
the same way that the cube root “undoes” what a cube “does.” This will allow us to find
unknown angles, given two sides of a right triangle.
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EXAMPLE:
EXAMPLE:
Find the length of the side labeled x.
Find the measure of the angle labeled θ .
x
7
35˚
5
15
SOLUTION:
cos 35 ο =
5
x
SOLUTION:
x=
5
≈ 6.104
cos 35 ο
tan θ =
7
15
θ = tan −1
7
≈ 25.017 ο
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EXERCISES:
1. In each of the right triangles below, find sin A, cos A, tan A, sin B, cos B,
and tan B.
12
(a) C
(b)
A
A
5
13
2
B
B
3
1
C
2. Find all unknown sides and angles for each of the following triangles.
This is called solving the triangle.
(a)
(b)
27˚
8
15
6
6
(c)
(d)
2
31
53˚
42˚
III. Applications
The following problem solving approach will be helpful in solving the many application
problems of right triangle trigonometry:
Sketch a picture.
Label your sketch with all known sides and angles.
Define variables for the unknown sides and angles.
Use the Pythagorean Theorem and the trigonometric ratios to find unknown
measures.
Before we jump into some problems, we should discuss angles of elevation and angles of
depression. Suppose you see an owl perched on the branch of a tree. The angle formed
between the ground and your line of sight to the owl is called an angle of elevation.
When flying to Hawai’i, you can see the Big Island in the distance. The angle formed
between the horizontal line the plane is traveling on and the line of sight down to the
island is called an angle of depression.
angle of depression
angle of elevation
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EXAMPLE:
A 100-foot fire truck ladder is leaning against a wall. Find the distance the ladder goes
up the wall if it makes an angle of 43˚ with the ground.
SOLUTION:
First we make a helpful sketch and
label the known and unknown.
Now, sin 43ο =
100 ft
x
.
100
x
43˚
ο
Thus x = 100 sin 43 ≈ 68.2 feet .
EXAMPLE:
A surveyor is standing 42 m from the base of a redwood tree. She measures the angle of
elevation to the top of the tree as 33˚. Find the height of the tree.
SOLUTION:
First we make and label a sketch.
Now, tan 33ο =
x
x
.
42
Thus x = 42 tan 33ο ≈ 27.28 m tall.
33˚
42 m
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EXAMPLE:
A guy wire is anchored to the ground 20 ft from the base of a telephone pole. If it is
attached to the telephone pole 30 ft above the ground, find the angle made by the guy
wire and the ground.
SOLUTION:
As usual, we first make a sketch.
Now, tan α =
30
.
20
Thus α = tan −1
30
≈ 56.31ο.
20
30 ft
20 ft
EXERCISES
1. The most powerful lighthouse is on the coast of Brittany, France, and is 50
meters tall. Suppose you are in a boat just off the coast. How far from the
base of the lighthouse are you if your angle of elevation to the light source
is 12˚?
2. A hot air balloon takes off from the ground and floats along an open field.
If the angle of elevation from the initial take-off spot is 68˚ and the
balloon is 350 feet in the air, what is the balloon’s distance from its takeoff spot?
3. A peregrine falcon perched atop a tall building spots its lunch on the
ground below. If the prey is 1000 m from the base of the building, and the
building is 200 m tall, what is the angle of depression from the falcon to
the prey?
4. The sonar of a navy cruiser detects a submarine that is 4000 feet from the
cruiser. The angle between the water line and the submarine is 34˚. How
deep is the submarine?
5. An aerial photographer is in an airplane at an altitude of 10 km and sees
two towns directly east of the plane. The angles of depression to the
towns are 25˚ and 60˚. How far apart are the two towns?
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6. Thales, the first of the Seven Wise Men, is said to have computed the
height of the Great Pyramid of Cheops. If we were a given distance away
and knew our angle of elevation to the top, this would be a simple
trigonometric problem. This is not how Thales determined the height.
Use the internet to search out the method used by Thales.
IV. The Law of Sines and the Law of Cosines
What if we do not have a right triangle? Let us turn our attention to two other types of
triangles, acute and obtuse. Triangles with no right angle are referred to as oblique.
An acute triangle is one in which all of the
angles are less than 90˚.
An obtuse triangle is one in which there is
an angle greater than 90˚.
To solve oblique triangles, we will need to know the measure of at least one side and any
two other parts of the triangle. Here are the four possible cases:
1.
2.
3.
4.
Two angles and any side (AAS or ASA)
Two sides and an angle opposite one of them (SSA, the ambiguous case)
Three sides (SSS)
Two sides and their included angle (SAS)
The first two cases can be solved using the Law of Sines and the second two cases can be
solved using the Law of Cosines.
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The Law of Sines
If ABC is a triangle with sides a, b, and c, then
C
C
b
a
b
c
=
=
.
sin a sin b sin c
a
A
a
b
B
c
A
A is acute
B
c
A is obtuse
EXAMPLE (AAS):
A
ο
ο
For a triangle with C = 130 , B = 27 , and b = 30 ,
find the remaining angle and sides.
30
SOULUTION:
C 130˚
First we make a sketch.
ο
ο
a
ο
c
27˚
ο
Note, A = 180 − (130 + 27 ) = 23 .
B
By the Law of Sines,
Also,
a
30
=
ο
sin 23
sin 27 ο
c
30
=
ο
sin 130
sin 27 ο
c=
a=
ο
30 sin 23
≈ 25.82.
sin 27 ο
30 sin 130 ο
≈ 50.62.
sin 27 ο
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EXAMPLE (ASA):
For a triangle with A = 75 ο , B = 45 ο , and c = 10 ,
find the measure of side a.
SOLUTION:
A
75˚
10
b
ο
First note that C = 60 .
Using the Law of Sines, we have
a
10
=
.
ο
sin 75
sin 60 ο
Thus a =
B
45˚
a
C
10 sin 75 ο
≈ 11.15 .
sin 60 ο
EXAMPLE (SSA, the ambiguous case):
Consider the triangle with B = 55 ο , a = 8.5, and b = 7.3 .
Find the measure of angle A
SOLUTION:
It is unclear whether this is an acute triangle or an obtuse triangle (or if it is not possible).
We must sketch both:
C
C
8.5
B
Now,
55˚
8 .5
7 .3
=
sin A sin 55 ο
7.3
A
sin A =
8.5
B
55˚
7.3
A
8.5 sin 55 ο
≈ 0.9538 .
7 .3
This gives two angles A1 ≈ 72.52 ο and A2 ≈ 180 ο − 72.52 ο ≈ 107.48 ο .
Both of these angles give triangles, as we do not exceed 180˚.
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Let us first take the acute case, A1 ≈ 72.52 ο . Then C ≈ 52.48 ο . Using the Law of Sines,
we find that the third side is c ≈
7.3 sin 52.48 ο
≈ 7.0682 .
sin 55 ο
Now we look at the obtuse case, A2 ≈ 107.48 ο . This gives C ≈ 17.52 ο . Again using the
7.3 sin 17.52 ο
≈ 2.6828 .
Law of Sines, we have c ≈
sin 55 ο
Note: When the ambiguous case arises, it is possible to obtain two triangles (as above),
one triangle, or no triangle. How do you know which? Just work the problem under the
assumption that you will obtain two triangles, and the mathematics will reveal the truth.
If we have either SSS or SAS, we must use the Law of Cosines as a first step to solving a
non-right triangle. After this, we will finish with the Law of Sines, as it is a little easier
to apply.
The Law of Cosines
a 2 = b 2 + c 2 − 2bc cos A
If ABC is a triangle with sides a, b, and c, then b 2 = a 2 + c 2 − 2ac cos B .
c 2 = a 2 + b 2 − 2ab cos C
Note that the Law of Cosines will be used to either find a side or it’s opposite angle.
EXAMPLE:
A ship travels 60 miles due east, then adjusts its course northward. After traveling 80
miles in that direction, the ship is 139 miles from its point of departure. Describe the
change in bearing from point B to point C.
SOLUTION:
C
139 mi
A
60 mi
80 mi
B
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Since we do not know any of the three angles of this triangle, we must use the Law of
Cosines.
Now, 139 2 = 60 2 + 80 2 − 2(60)(80) cos B .
This gives cos B =
60 2 + 80 2 − 139 2
≈ −0.97094
2(60)(80)
B ≈ cos −1 (−0.97094) ≈ 166.15 ο .
So the bearing measured from due north from point B to point C is given by
166.15 ο − 90 ο = 76.15 ο . We write this as N 76.15 ο E, read 76.15 ο east of north.
EXERCISES
1. Sketch triangles for each of the following and solve each triangle.
(a) a = 6, b = 8, c = 10
(b) A = 30 ο , c = 40, a = 20
(c) A = 115 ο , b = 15, c = 10
(d) B = 56.3ο , a = 8.3, b = 7.6
(e) a = 55, b = 25, c = 72
(f) A = 85 ο , a = 15, b = 25
(g) C = 145 ο , b = 4, c = 14
(h) A = 24.3ο , C = 54.6 ο , c = 21.6
2. A ranger located at station A spots a fire in the direction 32˚ east of north.
Another ranger, located at station B, 10 miles due east of station A, spots
the same fire on a line 48˚ west of north. Find the distance from each
ranger station to the fire.
3. A surveyor is attempting to find the distance between two points A and B.
A grove of trees is obstructing the view, so the surveyor sets a stake on
each side of the grove at the points A and B and then moves to a point C.
The distance from A to C is 143 feet and the distance from B to C is 123
feet. Lastly, angle ACB measures 78˚. Find the distance across the grove.
4. Trigonometry is used extensively in aerial photography. Suppose a
camera lens has an angular coverage of 75˚. As a picture is taken over
level ground, the airplane’s distance is 4800 feet from a house located on
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the edge of the photograph and the angle of elevation of the airplane from
the house is 48˚. Find the distance across the photograph.
5. A 10-meter telephone pole casts a 17-meter shadow directly down a slope
when the angle of elevation of the sun is 42˚. Find the angle of elevation
of the ground.
6. Because of prevailing winds, a tree grew so that it was leaning 4˚ from the
vertical. At a point 35 meters from the tree, the angle of elevation to the
top of the tree is 23˚. Find the height of the tree.
7. Suppose you are in a hot air balloon with your crew on the ground.
Attached to the balloon are two tether cords of length 200 ft and 240 ft,
which your crew has attached to the ground. You note that these cords
form an angle of 65˚ where they meet the balloon. Assuming these lines
are taut and the ground is level, with what angles do these cords meet the
ground?
8. To approximate the length of a marsh, a surveyor walks 250 meters from
point A to point B, then turns 75˚ and walks 220 meters to point C.
Approximate the length AC of the marsh.
9. A 100-foot vertical tower is to be erected on the side of a hill that makes a
6˚ angle with the horizontal. Find the length of each of the two guy wires
that will be anchored 75 feet uphill and downhill from the base of the
tower.
10. The baseball player in center field is playing approximately 330 feet from
the television camera that is behind home plate. A batter hits a fly ball
that goes to the wall 420 feet from the camera. Approximate the number
of feet that the center fielder has to run to make the catch if the camera
turns 8˚ to follow the play.
(Note: Be very careful sketching pictures for these problems. There may be more than
one interpretation for a couple of them; solve them all!)
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