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Algebra 113
First Year of Algebra
Supplementary Unit on Trigonometry
Written by:
Wally Dodge
Pat Wadecki
Kathy Anderson
Becky Lee
David Fickett
Jean Hostetler
(modified)
First Year Algebra 113
Supplemental Trigonometry Section #1
In this section you will learn the following:
A) The definition of the 3 trigonometry ratios: sine (sin), cosine (cos), and tangent (tan)
B) How to use your calculator to find trigonometric ratios
C) How to use trigonometry to solve problems.
Consider the following right triangle.
hypotenuse
leg opposite angle 

leg adjacent to angle 
In mathematics we often use the Greek letter , pronounced “theta,” to indicate an angle in a
triangle. The leg that forms one of the sides of  is called the leg adjacent to . The other
leg is called the leg opposite from .
Example 1: Complete the chart for each triangle given below.
8
III

I
5
3
15

4
12

5
13
II
Table #1
Triangle
Leg
opposite 
Leg
adjacent 
hypotenuse
Ratio of
opposite leg
to hypotenuse
Ratio of
adjacent leg
to hypotenuse
Ratio of
opposite leg
to adjacent
leg
I
II
III
1
Solution to example 1 from previous page:
Table #1
Triangle
Leg
opposite 
Leg
adjacent 
hypotenuse
Ratio of
opposite leg to
hypotenuse
Ratio of adjacent
leg to
hypotenuse
Ratio of
opposite leg to
adjacent leg
I
3
4
5
3/5 = 0.6
4/5 = 0.8
3/4 = 0.75
II
5
12
13
5/13 ≈ 0.385
12/13 ≈ 0.923
5/12 ≈ 0.417
III
15
8
17*
15/17 ≈ 0.882
8/17 ≈ 0.471
15/8 = 1.875
*Note: We used the Pythagorean Theorem for triangle III, in order to find that the hypotenuse was 17.
Using a protractor to measure the angles marked  in each of the triangles labeled I, II, and III on
the previous page, we found the information summarized in the table below.
Table #2
Triangle
Measure of 
in degrees
I
≈ 36.9°
II
≈ 22.6°
III
≈ 61.9°
Example 2: Set your calculator into “Degree” mode and complete the following table by
choosing the appropriate keys on your calculator.
Table #3
Triangle
I
in
degrees
36.9°
II
22.6°
III
61.9°
sin()
cos()
tan()
Solution to example 2 from above:
Table #3
sin()
cos()
tan()
Triangle
I
in
degrees
36.9°
≈ 0.600
≈ 0.800
≈ 0.751
II
22.6°
≈ 0.384
≈ 0.923
≈ 0.416
III
61.9°
≈ 0.882
≈ 0.471
≈ 1.873
If you compare the columns with the ratios in Table #1 with the sin(), cos(), and tan()
columns in Table #3, you will notice something very interesting. The results are nearly
identical. This not a coincidence: these ratios are the definitions of sin(), cos(), and tan().
2
Definition: In a right triangle with an acute angle labeled , the trigonometric ratios sine(),
cosine(), and tangent() are defined as follows:
B
hypotenuse
leg opposite angle 

A
C
leg adjacent to angle 
sine(), abbreviated sin() =
leg opposite angle 
, or
hypotenuse
cosine(), abbreviated cos() =
leg adjacent angle 
, or
hypotenuse
tangent(), abbreviated tan() =
leg opposite angle 
, or
leg adjacent angle 
sin(A) =
BC
AB
cos(A) =
AC
AB
tan(A) =
BC
AC
Example 3: Given the figure below, find sin(M ), cos(M), and tan(M)
P
12
N
9
M
Solution to Example 3:
By the Pythagorean Theorem, MP = 15
P
sin(M) =
12
N
opposite leg PN 12 4


 = 0.8
hypotenuse PM 15 5
adjacent leg MN 9 3
cos(M) =

  = 0.6
hypotenuse PM 15 5
9
15
M
tan(M) =
opposite leg PN 12 4


 ≈ 1.333
adjacent leg MN 9 3
3
Example 4: For the figure given below of a right triangle with an acute angle of 35°, do the
following:
a) Using your knowledge of trigonometric ratios, write two equations, one involving x and
the other involving y.
b) Solve the equations that your wrote in part (a).
y
x
35°
4
Solution to Example 4:
a) tan(35°) =
x
4
and
cos(35°) =
4
y
Note: if you use the trigonometric ratio for sine you get sin(35°) =
x
which is an equation with both x and
y
y. It is a correct equation, but, since it has both variables in it, this equation isn’t useful to us.
b) Using the trig keys on the calculator (make sure you’re still in “degree” mode!), we obtain:
tan(35°) =
0.7002075 ≈
x
4
x
4
0.7002075*4 ≈ x
0.7002075*4 ≈ x
2.80 ≈ x
cos(35°) =
4
y
.819152 ≈
4
y
.819152*y ≈ 4
y≈
4
.819152
y ≈ 4.88
Practice Problems Using Trigonometry (solutions follow on pages 8 and 9)
1. Find the height of the flagpole given the information shown on the diagram.
51°
4
36 feet
2. An airplane is 50 miles from an airport and the angle of depression of the airport is 7°. How
high is the airplane above the ground?
airplane
7°
50 miles
airport
3. A picture that is 4 feet wide is to be hung from a wire making an angle of 40° with the
picture, as shown below. How long is the wire?
wire
●
Nail
40°
40°
4 feet
4. The Marina is directly across Springstead Lake from the Frontier Inn. To get from one to the
other without getting your feet wet, you measure horizontally 4575 feet from the Frontier Inn
to point A, and find that the measure of A is 25°. What is the direct distance, across the
lake, from the Marina to the Frontier Inn?
Frontier Inn
4575 feet
25°
Marina
A
5
SOLUTIONS to Section #1: Practice Problems Using Trigonometry
#1) Let x = the height of the flagpole, in feet. From the diagram, we have:
tan(51°) =
x
36
x = 36 * tan(51°)
calculator gives
x ≈ 44.46
So the height of the flagpole is approximately 44.46 feet.
#2) The angle of depression is the angle measured from the horizontal to the direction looking
downward, as shown in the diagram. (Further explanation on the angle of depression is
given on page 9.) Let x = the height of the airplane. From the diagram, we have:
sin(7°) =
x
50
x = 50 * sin(7°)
calculator gives
x ≈ 6.095
The airplane is approximately 6.095 miles above ground, which is a little over 32,000 feet.
(This is a reasonable cruising altitude for a commercial airliner.)
#3) To use trigonometry, we need a right triangle, so we draw in the height of the triangle formed
by the wire. This height divides the horizontal side of the picture into two equal pieces. Let
x = one-half the length of the wire, as shown in the diagram below.
From this diagram (with the height drawn in), we have:
x
40°
cos(40°) =
2
x
2
x * cos(40°) = 2
x=
2
cos(40  )
≈ 2.61
4 feet
Since x represents half the length of the wire, the wire is approximately 5.22 feet long.
6
More SOLUTIONS to Section #1: Practice Problems Using Trigonometry
#4) Let x = the distance, in feet, across the lake from the Marina to the Inn, as shown below.
From this diagram, we have:
tan(25°) =
Frontier Inn
4575 feet
25°
x
4575
A
x
x = 4575 * tan(25°)
x ≈ 2133.36
Marina
The distance across the lake from the Marina to the Frontier Inn is approximately 2133 feet.
NOTE:
For some applications of trigonometry, you need to know the meanings of angle of
elevation and angle of depression.
A
If an observer at point P looks upward
toward an object at A, the angle the line of
sight 
 makes with the horizontal
PA

 is called the angle of elevation.
PH
angle of elevation
P
H
P
If an observer at point P looks downward
toward an object at B, the angle the line of
sight 
 makes with the horizontal
H
angle of depression
PB

 is called the angle of depression.
PH
B
(from: Rhoad, Milauskus, & Whipple: Geometry for Enjoyment and Challenge, New Edition. McDougal, Little, & Company: 1991)
In class Example:
1. Tom is looking up at his friend, Jen who is in a tree. He estimates his angle of elevation to
be 40⁰ and is standing 20 feet away from the base of the tree. How high off the ground is
Jen, to the nearest foot?
2. Jack is looking out the window at Jill. He estimates his angle of depression to be 30⁰. He
know he’s about 45 feet above ground level. How far is Jill from the building?
7
Exercises for Trigonometry: Section I
B
1) a) Label the sides of the triangle opposite, hypotenuse,
and adjacent using A as the given angle.
b) Label the sides of the triangle opposite, hypotenuse,
and adjacent using B as the given angle.
A
2) For the given triangles, with angle , do the following:
i) Give the lengths of the side adjacent to , the side opposite , and the hypotenuse.
ii) Give the values of sin(), cos(), and tan().
4
a)
b)
13
3
5


10
c)
d)
6

45
12

3) Find sin(A), cos(A), and tan(A) in terms of a, b, and c.
a
B
b
c
A
8
4) Using your knowledge of trigonometry, find the length of the missing side marked x on each
figure.
x
a)
°
b)
15
°
10
x
c)
d)
°
x
x
°
14
12
5) Find the value of x in each case.
a)
b)
40°
c)
x
25°
3
x
7
45°
4
x
6) Given a right triangle ABC with the right angle at C.
a) Draw a possible picture of this right triangle.
b) If sin(A) = 3/5, then cos(A) =
c) If tan(A) = 2/3, then sin(A) =
(info from previous parts no longer true)
d) If cos(A) = 4/7, then tan(A) =
(info from previous parts no longer true)
7) Round your answers to this problem to the nearest hundredth. (Use the triangle shown!)
60°
2
1
30°
3
Find: a) sin(30°)
b) cos(60°)
c) sin(60°)
d) cos(30°)
e) tan(30°)
f) tan(60°)
9
8) Round your answers to the nearest hundredth.
5
5
45°
5 2
Find: a) sin(45°)
b) cos(45°)
c) tan(45°)
d) Why are sin(45°) and cos(45°) the same?
9) You are flying a kite on a string that is 150 feet in length. The angle that the kite string
makes with the ground is 75°. Find the height of the kite above the ground.
10) You are parasailing behind a boat. You are 125 feet above the water, and the rope that you
are holding (attached to the boat) makes an angle of 50° with the water. How long is the
rope that you are holding?
11) To find the distance across Lake Springstead without getting his feet wet, Sam the surveyor
measures the distance (3600 feet) and the angle (22°) shown in the diagram below. If B is
a right angle, how far is it across Lake Springstead?
3600 feet
22°
A
B
12) You are 60 feet due west from the front of the Eurostar train in Waterloo station, which is
facing North/South. You calculate an angle of 60° between the line from where you are
standing to the front of the train and the line from where you are standing to the rear of the
train. How long is the train?
10
13) You are at an elevation of 18,500 feet and you spot the peak
of Mt. McKinley about one mile away along your line of sight.
You can’t remember how high Mt. McKinley is, so you take out
some of your surveying equipment to find the angle of elev.
You find that the angle of elevation of the peak of Mt.McKinley,
from where you are standing, is approximately 20.2°.
How high is this mountain peak?
Mt. McKinley
14) A guard standing on the lookout on top of a castle 250 feet high has spotted an army of
knights approaching at an angle of depression of 2°. How far is this army from the castle?
15) Heimlich maneuvered his mountain bike straight up the side of a hill with a 16° grade. If the
hill had an elevation (= height) of 1600 feet, how long was the trail up the hill?
16) Mario parked his car at a 45° angle with the curb,
as shown in the diagram. If the car is 14 feet long
and 6 feet wide, how far is the right rear corner of
the car from curb? How far is the left front corner
from the curb?
car
45°
right rear corner
17) Mrs. Bluebird is on a mad hunt for a worm to give to her babies for breakfast. She is flying
at an altitude of 34 feet when she spots a worm at an angle of depression of 19°. How far
must she travel to catch breakfast for her babies?
18) Rectangle ABCD is inscribed in circle O. (That is, it just fits inside the circle, and it’s
corners are labeled A, B, C, D as you walk around the circle.) If the radius of circle O is 19
and the measure of BDC is 30°, find the perimeter of triangle ABD.
19) A satellite hovers 300 km above the planet. The angle from the satellite to the horizon is 80°,
as shown in the diagram. What is the radius of the planet?
80°
(satellite)
●
P
300 km
11
First Year Algebra 113
Supplemental Trigonometry Section #2
In this section you will learn the following:
A) How to find an angle when given a trigonometric ratio
B) How to use your calculator to find angles
C) How to use trigonometry to do application problems.
In the last section you learned how to find lengths of sides in a right triangle when given one side
and an acute angle in the triangle. In this section you will learn how to find an acute angle in a
right triangle when given two sides of that triangle.
Use your graphing calculator’s “table” tool to find an approximate value for the
angle shown in each triangle below.
Example 1:
I
II
4

5
6

7
Solution
For Triangle I: tan() =
4
≈ 0.5714; in “degree” mode, let y1 = tan(x) and set up a
7
table to start at 0° and move in steps of 1°, as shown below. To the nearest degree,  ≈ 30°.
For Triangle II: sin() =
5
≈ 0.8333; this time, let y1 = sin(x) and again set up a table to
6
start at 0° and move in steps of 1°, as shown above. To the nearest degree,  ≈ 56°.
Earlier this year, we discussed inverse operations. We learned, for example, that the inverse
operation for dividing is to multiply, and that the inverse operation for square root is to square.
What we need is a function that would do the inverse operation of a trigonometric function. That
is, suppose that sin() =
to find the value of ?
5
. What operation could we do to both sides of the equation in order
6
Fortunately, there is an inverse operation for sine, called the inverse sine function. This
function is illustrated in the following example.
12
5
6
Example 2: Suppose sin() = . What is the size of the angle , in degrees?
Solution: The inverse of the sine function is the labeled “sin−1” on your calculator – it is located
above the “sin” key. In order to find the angle  we do the following:
sin() =
5
6
5
6
sin−1(sin()) = sin−1( )
5
6
Since sin−1 is the inverse operation for sine we obtain  = sin−1( ). We use a calculator to
5
6
find that sin−1( )≈ 56.44°.
Example 3: Nate is watching a ball game from the top of Comiskey Park. His seat is 130 feet
above the playing field and his distance from the pitcher is 160 feet. At what angle does he
need to tilt his head down in order to see the pitcher?
Solution: Below is a drawing to describe the situation

Nate
130 ft
160 ft
pitcher
We see the following trigonometric ratio: sin() =
130
. Using our inverse operation for
160
sine, we obtain
sin−1(sin()) = sin−1(
 = sin−1(
130
)
160
130
) ≈ 54.34°
160
13
Although we have only used the inverse operation for sine, namely sin−1, there are also inverse
functions for both the cosine and tangent. We formalize this in the following definition:
Definition:
a) The inverse operation for sine, called the inverse sine function, is denoted by sin−1.
b) The inverse operation for cosine, called the inverse cosine function, is denoted by cos−1.
c) The inverse operation for tangent, called the inverse tangent function, is denoted by tan−1.
Practice Problems Using Inverse Trigonometric Functions (solutions follow on p. 18 and 19)
1. Find the measures of A and C in the right triangle given below:
12
A
B
5
C
2. Find the value of x and the value of  in the figure below.
x

40°
12
9
3. If you are standing at the origin of a rectangular coordinate system, and are facing the
positive x-axis, by what angle do you have to turn in order to look directly at the ordered
pair (5, 15) ?
y
● (5, 15)
x
14
SOLUTIONS to Section II: Practice Problems Using Inverse Trigonometric Functions
#1)
12
A
B
5
C
From the diagram:
tan(A) =
5
12
tan−1(tan(A)) = tan−1(
A = tan−1(
tan(C) =
5
)
12
12
5
tan−1(tan(C)) = tan−1(
5
)
12
C = tan−1(
A ≈ 22.6°
12
)
5
12
)
5
C ≈ 67.4°
Note: Once you found that A ≈ 22.6°, if you remember that all angles in a triangle add up to
180°, you can find C by subtraction: C = 180 − 90 − 22.6 = 67.4°.
#2)
x

40°
12
9
Using the right triangle at the left of the picture, we have:
tan(40°) =
x
.
9
Therefore 9∙tan(40°) = x, and thus x ≈ 7.5519
Now using this result and the right triangle at the right of the picture we obtain:
x
7.5519
≈
12
12
7.5519
tan−1(tan()) = tan−1(
)
12
tan() =
 ≈ 32.18°
Note: Be careful about rounding error….. see how we used four decimals in our approximation
of x when we used it to find . Better yet, use your calculator’s “ANS” key!
15
MORE solutions to Section II: Practice Problems Using Inverse Trigonometric Functions
#3) We first add some lines to the original diagram to obtain:
y
● (5, 15)
tan() =
15
=3
5
tan−1(tan()) = tan−1(3)


 ≈ 71.57°

NOTE
x
The sum of the angles in a triangle is:
Isosceles Triangles:
Exercises for Trigonometry: Section II
1. Find the missing angle labeled A in each diagram.
a) A
b)
5
3

5
3
c)
d)

3
3
4
7

16
2. Find the values of x and y in the diagram below.
12
24
x
y
40°
3. Find the values of x and y in the diagram at right.
(Both x and y are labeling angles.)
6.5
4
y
x
4
4
4. An isosceles triangle ABC has two equal sides of 14
inches long, and the base is 8 inches long. Find the measures
of A, C, and ABC (in degrees).
B
14
14
A
C
8
5. You are standing at the origin of a rectangular coordinate system, facing out towards the
positive x-axis. How many degrees do you rotate counter-clockwise to be facing directly at
the point (2, 21) ?
6. From the right bank of the Seine, you spot the top of the 985 foot high Eiffel Tower, one mile
in the distance. Find the angle of elevation to the top of the tower.
7. You are flying at an altitude of 3300 feet and spot the London Bridge, 5000 feet from where
you are. Find the angle of depression from you to the bridge.
8. Dubious wanted to make a skateboard ramp by nailing
plywood over 4 stairs on his family’s backyard deck.
Each stair had a run (depth) of 10 inches and a rise of 8 inches.
What would be the angle of elevation of the ramp?
ramp
8
10
17
9. In right ABC with right angle at C, the length AC = 4.32 cm. and the length AB = 8.4 cm.
Find the measure of B, rounded to the nearest tenth of a degree.
10. A destroyer in the Atlantic Ocean is 2300 feet above the ocean’s floor when the captain
notices an enemy submarine sitting on the ocean floor, 6200 feet away. At what angle of
depression must the destroyer aim its missile to hit this enemy submarine?
11. ABC is a right triangle with right angle at vertex C. If the cosine of A is
measure of B.
5
, find the
13
12. You are driving up a hill that is 2.8 miles long and 902.4 feet high. Find the angle of
elevation of the hill.
13. For the figure pictured at right, find:
a. the perimeter of the rectangle.
132”
b. the area of the rectangle.
25°
14. To find the height of the smokestack at New Trier, a student takes some measurements. The
distance from where he is standing to the bottom of the smokestack is 110 feet, and, using a
clinometer, he finds that the angle of elevation to the smokestack is 36.2°. How tall is the
smokestack?
15. Find the angle  in the diagram.
350 feet

23°
1400 feet
18
16.
A
B
C
15
F
13
G
6
E
38°
D
For the figure given above, with rectangles ACDE and ABGF, find each of the following:
a) the area of rectangle ABGF
b) the length of DE
c) the area of rectangle ACDE
d) the area of trapezoid FGDE
e) the area of trapezoid BCDG
17. A tent is in the shape of a cone with a diameter of 14 feet. The peak rises 8 feet above the
center of the floor. How far from the edge must a 5-foot tall person stand in order to be able
just barely touch the tent fabric with her head?
8
5
14
19
ANSWERS: Section I
1) a)
b)
opposite
adjacent
B
B
adjacent
opposite
hypotenuse
hypotenuse
A
2) a)
b)
c)
d)
i)
i)
i)
i)
A
adjacent = 3, opposite = 4, hypotenuse = 5
adjacent = 12, opposite = 5, hypotenuse = 13
adjacent = 12, opp = 6, hypotenuse ≈ 13.42
adjacent ≈ 43.87, opp = 10, hypotenuse = 45
ii)
ii)
ii)
ii)
sin() = 4/5, cos() = 3/5, tan() = 4/3
sin() = 5/13, cos() = 12/13, tan() = 5/12
sin() ≈ .4472, cos() ≈ .8944, tan() = .5
sin() = 2/9, cos() ≈ .9750, tan() ≈ .2279
3) sin() = a/c, cos() = b/c, tan() = a/b
4) a) ≈ 4.38
b) ≈ 12.29
c) ≈ 4.37
5) a) ≈ 2.52
b) ≈ 2.96
c) ≈ 5.66
6) a)
d) ≈ 38.46
6c) ≈ .5547
6b) 4/5
B
C
6d) ≈ 1.4361
A
7) a) 1/2
b) 1/2
c) 0.87
d) 0.87
e) 0.58
f) 1.73
8) a) 0.71
b) 0.71
c) 1
d) an isosceles triangle has two equal sides, so the
opposite and adjacent are the same
9) ≈ 144.89 feet
10) ≈ 163.18 feet
11) ≈ 1348.58 feet
12) ≈ 103.92 feet
13) ≈ 20323 feet
14) ≈ 7159.06 feet
15) ≈ 5804.73 feet
16) right rear to curb ≈ 9.9 feet; left front to curve ≈ 4.24 feet
17) ≈ 104.43 feet
18) perim ≈ 89.91
19) ≈ 19446.9 kilometers
Section II
1) a) ≈ 30.96°
b) ≈ 36.87°
2) x ≈ 7.71, y ≈ 18.75°
c) ≈ 55.15°
d) 45°
3) x ≈ 35.66°, y ≈ 71.32°
4) A = C ≈ 73.40°; ABC ≈ 33.20°
5) ≈ 84.56°
7) ≈ 41.30°
8) ≈ 38.66°
10) ≈ 21.78°
12) ≈ 3.5°
13) a) ≈ 350.84 units
14) ≈ 80.51 feet
16) a) 117
9) ≈ 30.9°
6) ≈ 10.75°
11) ≈ 22.62°
b) ≈ 6673.78 square units
15) ≈ 11°
b) ≈ 20.68
c) ≈ 310.19
d) ≈ 101.04
e) ≈ 92.16
17) 4.375 feet from the edge of the tent
20