Trigonometry Unit
(Level IV Academic Math)
NSSAL
(Draft)
C. David Pilmer
©2010
(Last Updated: Dec 2011)
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Acknowledgments
The Adult Education Division would also like to thank the following NSCC instructors for
piloting this resource and offering suggestions during its development.
Charles Bailey (IT Campus)
Elliott Churchill (Waterfront Campus)
Barbara Gillis (Burridge Campus)
Barbara Leck (Pictou Campus)
Suzette Lowe (Lunenburg Campus)
Floyd Porter (Strait Area Campus)
Brian Rhodenizer (Kingstec Campus)
Joan Ross (Annapolis Valley Campus)
Jeff Vroom (Truro Campus)
Table of Contents
Introduction …………………………………………………………………………………..
Negotiated Completion Date …………………………………………………………………
The Big Picture ………………………………………………………………………………
Course Timelines …………………………………………………………………………….
Prerequisite Knowledge ……………………………………………………………………...
ii
ii
iii
iv
v
The Pythagorean Theorem …………………………………………………………………...
Trigonometric Ratios ………………………………………………………………………...
Using Our Three Trigonometric Ratios, Part 1 ………………………………………………
Using Our Three Trigonometric Ratios, Part 2 ………………………………………………
The Law of Sines …………………………………………………………………………….
Using the Law of Sines ………………………………………………………………………
The Law of Cosines ………………………………………………………………………….
Using the Law of Cosines ……………………………………………………………………
Putting It Together …………………………………………………………………………...
1
7
11
17
22
25
34
37
49
Post Unit Reflection …………………………………………………………………………. 55
Which Formula Should I Use? ………………………………………………………………. 56
Answers ……………………………………………………………………………………… 57
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Introduction
Trigonometry is a branch of mathematics that studies
1. the relationships between sides and angles of triangles, and
2. trigonometric functions that result from real world situations that are periodic in nature (i.e.
repeat at regular intervals).
In this unit, we will focus on the relationships between the sides and angles of triangles. Initially
we will only work with right-angle triangles where we will use the Pythagorean Theorem and the
three trigonometric ratios (sine, cosine, and tangent) to solve a variety of problems. Later in the
unit, we will learn about the two laws (Law of Sines and the Law of Cosines) that can be applied
to find the missing sides and/or angles of any triangle (rather than be restricted to only rightangle triangles).
In the next unit, Sinusoidal Functions Unit, we will investigate the other component of
trigonometry, specifically trigonometric functions. We will examine a particular type of
trigonometric function called sinusoidal functions. These functions create curves that look like
waves.
Trigonometry in its most primitive forms was developed by the Greeks to assist in their study of
astronomy. Indian, Islamic and Chinese mathematicians made further advancements in the field
between the 6th and 16th centuries. Today trigonometry is used in many disciplines (surveying,
engineering, navigation, acoustics, astronomy, seismology, drafting).
Negotiated Completion Date
After working for a few days on this unit, sit down with your instructor and negotiate a
completion date for this unit.
Start Date:
_________________
Completion Date:
_________________
Instructor Signature: __________________________
Student Signature:
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The Big Picture
The following flow chart shows the optional bridging unit and the eight required units in Level
IV Academic Math. These have been presented in a suggested order.
Bridging Unit (Recommended)
• Solving Equations and Linear Functions
Describing Relations Unit
• Relations, Functions, Domain, Range, Intercepts, Symmetry
Systems of Equations Unit
• 2 by 2 Systems, Plane in 3-Space, 3 by 3 Systems
Trigonometry Unit
• Pythagorean Theorem, Trigonometric Ratios, Law of Sines,
Law of Cosines
Sinusoidal Functions Unit
• Periodic Functions, Sinusoidal Functions, Graphing Using
Transformations, Determining the Equation, Applications
Quadratic Functions Unit
• Graphing using Transformations, Determining the Equation,
Factoring, Solving Quadratic Equations, Vertex Formula,
Applications
Rational Expressions and Radicals Unit
• Operations with and Simplification of Radicals and Rational
Expressions
Exponential Functions and Logarithms Unit
• Graphing using Transformations, Determining the Equation,
Solving Exponential Equations, Laws of Logarithms, Solving
Logarithmic Equations, Applications
Inferential Statistics Unit
• Population, Sample, Standard Deviation, Normal Distribution,
Central Limit Theorem, Confidence Intervals
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Course Timelines
Academic Level IV Math is a two credit course within the Adult Learning Program. As a two
credit course, learners are expected to complete 200 hours of course material. Since most ALP
math classes meet for 6 hours each week, the course should be completed within 35 weeks. The
curriculum developers have worked diligently to ensure that the course can be completed within
this time span. Below you will find a chart containing the unit names and suggested completion
times. The hours listed are classroom hours. In an academic course, there is an expectation that
some work will be completed outside of regular class time.
Unit Name
Minimum
Completion Time
in Hours
0
6
18
18
20
36
12
20
20
Total: 150 hours
Bridging Unit (optional)
Describing Relations Unit
Systems of Equations Unit
Trigonometry Unit
Sinusoidal Functions Unit
Quadratic Functions Unit
Rational Expressions and Radicals Unit
Exponential Functions and Logarithms Unit
Inferential Statistics Unit
Maximum
Completion Time
in Hours
20
8
22
20
24
42
16
24
24
Total: 200 hours
As one can see, this course covers numerous topics and for this reason may seem daunting. You
can complete this course in a timely manner if you manage your time wisely, remain focused,
and seek assistance from your instructor when needed.
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Prerequisite Knowledge
Before getting deep into the Trigonometry Unit, we have to review a few concepts.
Naming Angles
The most common way to name an angle is to use the three letters
on the shape that define the angle, with the middle letter
representing the vertex of the angle. In the diagram
∠CAB = 40° , ∠ABC = 50° , and ∠BCA = 90° .
C
90o
40o
With simple geometric figures, as is the case with the diagram
provided, the vertex alone can be used to define the angle.
∠A = 40° , ∠B = 50° , and ∠C = 90°
A
Naming Sides of Triangles
The side opposite (across from) a specific vertex on a triangle is
named using the same letter as the vertex but using a lower case
letter. In the diagram, the side opposite vertex A or ∠A is called
side a. The side opposite vertex B or ∠B is called side b. The side
opposite vertex C or ∠C is called side c.
Interior Angles of a Triangle
The interior angles of any triangle add up to 180o.
In the diagram:
∠ABC + ∠BAC + ∠ACB
= 48° + 25° + 107°
= 180°
Interior Angles of a Quadrilateral
The interior angles of any quadrilateral (i.e. four-sided figure)
add up to 360o.
In the diagram:
∠HEF + ∠EFG + ∠FGH + ∠GHE
= 100° + 128° + 61° + 71°
= 360°
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50o
B
C
b
A
a
c
B
C
A
B
H
G
E
F
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The Pythagorean Theorem
The Pythagorean Theorem, a formula that describes the relationship among the three sides of a
right-angle triangle, is probably one of most famous and useful theorems used in mathematics.
The theorem is named after the Greek mathematician, Pythagoras (570 BC to 495 BC), who is
credited for the proof, even though many argue that the actual theorem was used previously by
the Egyptians in their surveying.
B
The theorem states that for a right-angle triangle ABC where c is
the hypotenuse (i.e. the longest side or the side opposite the
c
right-angle), the relationship between the three sides, a, b, and c,
a
can be described by the formula a 2 + b 2 = c 2 . In words we are
saying that in a right-angled triangle the square of the hypotenuse
is equal to the sum of the squares of the other two sides (referred
A
b
C
to as the legs of the triangle). Notice the relationship between the
naming of the angles of the triangle (in capital letters) and the naming of the sides (in lower
case). The side opposite angle A (or ∠A ) is called a. The side opposite angle B is called b.
Does this formula actually work? Consider the triangle that has been drawn below. Al three
sides have been measured accurately to two decimal points. When these values are entered into
the formula a 2 + b 2 = c 2 , everything works out (i.e. both sides of the equation are equal to each
other.
a2 + b2 = c2
c = 7.66 cm
4.58 2 + 6.14 2 = 7.66 2
20.98 + 37.70 = 58.68
58.68 = 58.68
a = 4.58 cm
b = 6.14 cm
The example is not a formal proof. We don’t have enough time to go over Pythagoras’ proof, or
other proofs, however; if you search the internet, you can find these proofs.
Example 1
Determine the missing side of the triangle.
P
Answer:
In this case, we are given the two legs of the triangle
and must find the hypotenuse.
a2 + b2 = c2
- Use the variables provided.
r 2 + p2 = q2
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3.9 m
Q
2.3 m
q
R
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3.9 2 + 2.3 2 = q 2
15.21 + 5.29 = q 2
20.5 = q 2
20.5 = q
q = 4.5 m
Example 2
Determine the missing side of the triangle.
G
Answer:
In this case, we are given the one leg and the
hypotenuse, and must find the other leg.
a2 + b2 = c2
e2 + f 2 = g 2
6.9 2 + f 2 = 9.2 2
6.9 cm
E
F
9.2 cm
47.61 + f 2 = 84.64
f 2 = 84.64 − 47.61
f 2 = 37.03
f = 37.03
f = 6.1 cm
4.7 m
Example 3
Determine the length of x.
9.3 m
Answer:
This question requires that we use the
Pythagorean Theorem twice. We use it first to
find the side shared by both right-angle
triangles. Then the theorem is used to find x.
a2 + b2 = c2
a2 + b2 = c2
8.5 2 + b 2 = 9.3 2
4.7 2 + 3.8 2 = x 2
72.25 + b 2 = 86.49
22.09 + 14.44 = x 2
b 2 = 86.49 − 72.25
36.53 = x 2
b 2 = 14.24
36.53 = x
x = 6.0
b = 14.24 = 3.8
x
8.5 m
The length of x is 6.0 m.
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Example 4
Tom traveled directly north for 2.4 km and then headed directly west for 1.9 km. In the end,
how far was Tom from his starting point?
Answer:
In this case we have to create a diagram.
Once we have the diagram, we can see that
we have the two legs and we must find the
hypotenuse.
a2 + b2 = c2
1.9 km
N
W
?
2.4 km
E
S
1.9 2 + 2.4 2 = c 2
3.61 + 5.76 = c 2
Starting
Point
9.37 = c 2
9.37 = c
c = 3.1 km
Tom is 3.1 km from his starting point.
Questions
1. Solve for the missing side.
(a)
12.7 cm
5.9 cm
(b)
9.8 m
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16.7 m
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(c)
R
9.1 cm
11.7 cm
P
Q
(d)
46 ft
S
T
52 ft
U
2. Determine whether the following triangle is right-angled.
D
8m
17 m
E
15 m
F
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3. For each of the following diagrams, solve for x.
(a)
5
13
x
16
(b)
15.7
x
8.3
20.3
(c)
43
x
35
26
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4. Maurita leans a 7 metre ladder against a vertical brick wall. If the
base of the ladder is 1.5 metres from the brick wall, how far up
the wall is the other end of the ladder?
5. The local community centre has decided to install a new wheelchair ramp. They know that
the ramp will rise 1.4 metres for a run of 11.2 metres. How long will the ramp be?
rise
run
6. Nancy traveled directly south for 1.7 km and then headed directly east for 0.8 km. In the
end, how far was Nancy from her starting point?
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Trigonometric Ratios
In the previous section, we learned about a special relationship that exists between the sides of a
right angle triangle. This relationship is called the Pythagorean theorem, and it is described by
the equation a 2 + b 2 = c 2 . In this section, we will investigate if a relationship exists between the
sides and angles of right angle triangles. Before we can start the investigation, we have to
understand how to label the sides of the right angle triangle. The hypotenuse is the easiest side
to identify because it is the longest side and always opposite the right angle. The opposite and
adjacent sides can only be identified when we know the angle of interest.
In the first diagram, the angle of interest is represented
by the Greek symbol θ (theta). The side that is
opposite θ (i.e. across from θ ) is called the opposite
side. The side that is beside θ , but not the hypotenuse,
is called the adjacent side.
hypotenuse
opposite
θ
adjacent
In the second diagram, the angle of interest θ is in
another location. For this reason, the opposite and
adjacent sides are in new locations.
hypotenuse
θ
adjacent
opposite
Investigation:
Part 1
You have been given three similar right-angle triangles; ∆ABC, ∆ADE, and ∆AFG. All of these
triangles share the same base angle of 20o. Measure the sides of the three triangles accurately to
the nearest millimeter and record the information in the table below. Calculate the ratios in the
last three columns using the measurements you recorded. Round the answers to the nearest
hundredths. The first set of measurements and calculations have been completed for you.
F
D
B
A
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20o
C
E
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Triangle
(20o)
Opposite
Adjacent
Hypotenuse
Opposite
Hypotenuse
Adjacent
Hypotenuse
Opposite
Adjacent
∆ABC
19 mm
57 mm
60 mm
0.32
0.95
0.33
∆ADE
26 mm
74 mm
79 mm
∆AFG
What pattern or patterns to you see?
Part 2
You have been given three similar right-angle triangles; ∆ABC, ∆ADE, and ∆AFG. In this case
they all share the base angle of 30o. Measure the sides of the three triangles and record the
information. Also calculate the three ratios identified in the last three columns.
F
D
B
A
Triangle
(30o)
30o
C
Opposite
Adjacent
E
G
Hypotenuse
Opposite
Hypotenuse
Adjacent
Hypotenuse
Opposite
Adjacent
∆ABC
∆ADE
∆AFG
What pattern or patterns to you see? Did you see similar patterns in Part 1?
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Part 3
You have been given three similar right-angle triangles; ∆ABC, ∆ADE, and ∆AFG. In this case
they all share the base angle of 40o. Measure the sides of the three triangles and record the
information. Also calculate the three ratios identified in the last three columns.
F
D
B
40o
A
Triangle
(40o)
C
Opposite
Adjacent
E
G
Hypotenuse
Opposite
Hypotenuse
Adjacent
Hypotenuse
Opposite
Adjacent
∆ABC
∆ADE
∆AFG
What pattern or patterns to you see? Did you see similar patterns in Parts 1 and 2?
Multiple Choice Questions
1. What do all nine triangles in parts 1, 2, and 3 have in common? They are all:
(a) isosceles
(b) equilateral
(c) similar
(d) right-angle
2. What do all three triangles in part 1 have in common? With the exception of what you
identified in question 1, they are all:
(a) isosceles
(b) equilateral
(c) similar
(d) right-angle
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3. Make sure your calculator is in "Degree" mode. Using your calculator, find cos 20o (cosine
of 20 o). Based on this answer and the work you completed in part 1, one can conclude that
the cosine of an angle is equal to:
Opposite
Adjacent
Opposite
(a)
(b)
(c)
Adjacent
Hypotenuse
Hypotenuse
4. Using your calculator, find sin 30o (sine of 30 o). Based on this answer and the work you
completed in part 2, one can conclude that the sine of an angle is equal to:
Opposite
Opposite
Adjacent
(a)
(b)
(c)
Adjacent
Hypotenuse
Hypotenuse
5. Using your calculator, find tan 40o (tangent of 40 o). Based on this answer and the work you
completed in part 3, one can conclude that the tangent of an angle is equal to:
Opposite
Adjacent
Opposite
(a)
(b)
(c)
Hypotenuse
Hypotenuse
Adjacent
Conclusions:
In this investigation you discovered three trigonometric ratios that can be applied to right-angle
triangles. Complete the ratios using the words opposite, adjacent, and hypotenuse. In each
equation we have used the Greek symbol θ (pronounced theta) to represent our angle of interest.
Sine Ratio:
sin θ =
hypotenuse
Cosine Ratio:
Tangent Ratio:
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cos θ =
opposite
θ
adjacent
tan θ =
10
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Using Our Three Trigonometric Ratios, Part 1
In the last section we learned about three trigonometric ratios (sine, cosine and tangent) that can
be applied to right-angle triangles.
sin θ =
opp
hyp
cos θ =
adj
hyp
tan θ =
opp
adj
hyp
opp
θ
adj
Some learners struggle remembering these three formulas. Here are two common memory tricks
that you may wish to use.
1. Use the acronym SOH-CAH-TOA
2. Remember the saying “Some Officers Have Curly Auburn Hair Till Old Age”
We can use these formulas to solve for unknown sides or angles in rightangle triangles. To accomplish this, we will need to learn how to use
some important keys on your calculator. First we must make sure our
calculator is in “degree” mode (rather than radian mode). On a regular
scientific calculator, the screen needs to display in small print either D or
degree. On the TI-83/84 graphing calculator, press MODE and select
Degree.
We will be using the SIN, COS, TAN commands on the calculator when we have been supplied
with angle θ .
We will be using the SIN-1, COS-1, TAN-1 commands when we need to find the angle θ .
Example 1
Determine x.
8.6
x
25o
Answer:
Start by labeling the two sides that have been identified in
the question.
Since you are working with the opposite side and the
hypotenuse, we must use the sine ratio because it is the
only formula that works with both the opposite side and
hypotenuse.
opp
sin θ =
hyp
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x
opp
8.6
hyp
25o
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Now substitute the two known values (25o and 8.6) and the unknown variable (x) into the
equation.
x
sin 25° =
8.6
Making sure that your calculator is in degree mode, find the sine of
25o and substitute this value into the equation. On a scientific this is
accomplished by entering 25 and then pressing SIN. On a TI-83/84
graphing calculator this is accomplished by pressing SIN, entering
25, and then pressing ENTER.
x
0.423 =
8.6
Now cross-multiply and solve for x.
0.423 × 8.6 = x
3.6 = x
Example 2
Determine x.
23.8
67o
x
Answer:
Label the sides identified in the question.
Since we are dealing with the adjacent side and the
hypotenuse, we need to work with the cosine ratio.
adj
cos θ =
hyp
23.8
adj
67o
x
hyp
Substitute the known values and the unknown variable into the equation.
23.8
cos 67° =
x
Using a calculator find the cosine of 67o.
23.8
0.391 =
x
Cross-multiply and solve for x.
0.391x = 23.8
23.8
x=
0.391
x = 60.9
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Example 3
Determine θ .
13.8
17.5
θ
13.8
opp
Answer:
Label the sides.
Use the sine ratio.
opp
sin θ =
hyp
13.8
sin θ =
17.5
sin θ = 0.789
17.5
hyp
θ
We do not use the SIN button. Instead we use the SIN-1 button to
find θ . On a scientific calculator, enter 0.789 and press SIN-1 (2nd,
SIN). On the TI-83/84 graphing calculator, press SIN-1 (2nd, SIN),
enter 0.789, and press ENTER.
θ = 52°
Example 4
Determine θ .
8.3
θ
11.9
Answer:
Label the sides.
Use the tangent ratio.
opp
tan θ =
adj
8.3
tan θ =
11.9
tan θ = 0.697
θ = 35°
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8.3
opp
θ
11.9
adj
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Questions
1
(a) Solve for x.
x
(b) Solve for x.
9.8
27o
16.7
35o
x
(c) Solve for θ .
(d) Solve for θ .
22.7
θ
12.9
θ
9.2
5.7
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(f) Solve for θ .
(e) Solve for x.
x
θ
21o
24.7
10.3
31.5
(g) Solve for θ .
(h) Solve for x.
x
34.6
28.1
0.75
o
50
θ
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(j) Solve for θ .
(i) Solve for x.
θ
x
81.7
112.0
o
62
74.9
(l) Solve for θ .
(k) Solve for x.
x
61.9
θ
23.1
48o
9.4
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Using Our Three Trigonometric Ratios, Part 2
Many of the problems we will encounter will require us to use more than one trigonometric ratio
and/or incorporate the Pythagorean Theorem.
Example 1
Solve for x.
B
C
o
x
22
31o
A
94 m
E
176 m
D
Answer:
There are three steps to solving this question.
Step 1:
Solve for CE using ∆CDE
and the cosine ratio.
adj
cos θ =
hyp
176
cos 31° =
CE
176
0.857 =
CE
0.857CE = 176
176
CE =
0.857
CE = 205.4
Step 2:
Solve for BE using ∆BCE
and the sine ratio.
opp
sin θ =
hyp
BE
sin 22° =
205.4
BE
0.375 =
205.4
0.375 × 205.4 = BE
BE = 77.0
Step 3:
Solve for AB (x) using
∆ABE and the Pythagorean
Theorem.
a2 + b2 = c2
94 2 + 77 2 = x 2
8836 + 5929 = x 2
14765 = x 2
14765 = x
x = 121.5 metres
Example 2
Ryan wants to know the height of a tree on his lot. He stands 15
metres from the base of the tree and measures the angle of elevation
from the ground to the top of the tree using a clinometer. He
discovers that the angle of elevation is 53o. Determine the height of
the tree.
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Answer:
We will be assuming that the tree is vertical and therefore at a right-angle to the ground. For
this question we need to find the opposite side using the tangent ratio.
opp
tan θ =
adj
h
Tree
tan 53° =
15
height
h
(h)
1.327 =
15
53o
1.327 × 15 = h
h = 19.9 metres
15 m
Note:
In many trigonometric word problems we will
encounter the terms angle of elevation and
angle of depression. These two terms are
used to describe the number of degrees one is
above or below the horizontal.
Angle of Elevation
Angle of Depression
horizontal
Questions
1. Solve for x.
x
63o
7.2 m
2. Solve for θ .
θ
153 cm
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241 cm
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3. Solve for x.
68o
49.2 m
52o
x
46.5 m
4. Solve for x and θ .
9.2 km
62o
x
34o
θ
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5. Taralee leans an 8 metre ladder against a vertical brick wall. If the
base of the ladder is 1.2 metres from the brick wall, what angle does
the ladder make with the ground?
6. A video camera is mounted on the top edge of a 12.3
metre building. The camera faces another building
directly across the road. If the camera is tilted up at
30o to the horizontal, it views the top edge of this
other building. The bases of the buildings are 18
metres apart. How tall is the building directly across
the street?
7. Jorell needs to figure out the width of a river. He uses a
post located on the other side of the river and his
knowledge of trigonometric ratios to accomplish this.
He was able to take two measurements on his side of
the river. These measurements are recorded on the
diagram. Determine the width of the river.
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post
35o
150 m
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8. The Pyramid of Khufu, completed in 2560
BC, is the oldest and largest of the three
pyramids of Giza. It is a square based
pyramid where the length of the base side is
230.4 m and the angle of elevation from the
middle of the base side to the top of the
pyramid is 50.3o. Determine the height of
the pyramid.
50.3o
230.4 m
9. The angle of descent for most passenger
aircraft is 3o. If a plane is at an altitude of
5 miles, how far away from the airport
(along a horizontal path) should the
aircraft start its descent?
3o
10. A flag and flag pole have been placed on the top of a tower. When
one stands 40 m from the center of the base of the tower, the angle
of elevation to the top of the tower is 54o. From the same position,
the angle of elevation to the top of the flagpole is 56.5o. How tall is
the flagpole?
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The Law of Sines
The Pythagorean Theorem and the three trigonometric ratios (sine, cosine, and tangent) only
work with right angle triangles. What happens when we are not dealing with a right-angle
triangle? In the example below, we have such a case.
Example 1
Determine x.
13.8 m
x
47o
63o
Answer:
With our limited knowledge of trigonometry, we are
forced to take this triangle and divide it into two right
angle triangles. By doing this, we can now use our
trigonometric ratios.
We will solve for y using the triangle on the left and the
sine ratio. We can then use the triangle on the right and
the sine ratio to solve for x.
opp
hyp
y
sin 47° =
13.8
y
0.731 =
13.8
0.731 × 13.8 = y
y = 10.1 m
sin θ =
13.8 m
x
y
47o
63o
opp
hyp
10.1
sin 63° =
x
10.1
0.891 =
x
0.891x = 10.1
x = 11.3 m
sin θ =
There has to be a better way to handle questions involving triangles that are not right angled.
There is. We use two different laws: Law of Sines and Law of Cosines. In this section we will
introduce the Law of Sines.
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Deriving the Law of Sines
We are only going to look at half of the derivation. We will show the derivation for acute
triangles (i.e. triangles where no interior angles exceeds 90o), but omit the derivation for obtuse
triangles.
We will start by dividing our triangle ( ∆ABC ) into two right-angle triangles (just as we did in
Example 1).
A
Important Note:
Notice that the side opposite angle
A is called a. Similarly the side
opposite angle B is called b. This
labeling technique is used when
employing the Law of Sines or
Law of Cosines.
b
c
B
D
C
a
For ∆ABD
opp
sin θ =
hyp
AD
sin B =
c
c sin B = AD
For ∆ACD
opp
sin θ =
hyp
AD
sin C =
b
b sin C = AD
We can substitute one equation into the other
because they are both equal to AD.
c sin B = b sin C
b sin C
c=
sin B
c
b
=
sin C sin B
Divide both sides by sin B.
Divide both sides by sin C.
If we continue this process of drawing lines perpendicular to the other sides, we can obtain the
following.
a
b
c
=
=
sin A sin B sin C
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This is the Law of Sines.
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Checking the Law of Sines
Below you have a triangle. Using a ruler and a protractor, measure the three sides and the three
angles. Record the values in the chart. Also complete the three calculations in the chart.
A
B
a=
∠A =
b=
∠B =
c=
∠C =
a
=
sin A
b
=
sin B
c
=
sin C
C
What do you notice? Is this what you expected?
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Using the Law of Sines
In the previous section we were introduced to the Law of Sines.
a
b
c
=
=
sin A sin B sin C
B
c
a
A
C
b
Example 1
Determine x.
13.8 m
x
63o
47o
Answer:
In the previous section we handled this question by dividing it into two right-angle triangles.
We can avoid this time-consuming process by using the Law of Sines, which is not restricted
to right angle triangles.
a
b
=
sin A sin B
x
13.8
=
sin 47° sin 63°
x
13.8
=
0.731 0.891
13.8
x = 0.731 ×
0.891
x = 11.3 m
Example 2
Solve for ∠E .
G
14.3 m
F
o
40
11.7 m
E
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Answer:
e
g
=
sin E sin G
14.3
11.7
=
sin E sin 40°
14.3
11.7
=
sin E 0.643
14.3 × 0.643 = 11.7 × sin E
14.3 × 0.643
= sin E
11.7
sin E = 0.786
E = 52°
Example 3
Determine ∠C .
14.8 ft
B
A
15.6 ft
C
13.2 ft
72o
37o
102o
D
E
Answer:
It takes three steps to solve this question.
Step 1:
Step 2:
Using ∆ABE and the
Using ∆BDE and the Law of
Pythagorean Theorem, solve Sines, solve for BD.
for BE.
d
e
=
a2 + b2 = c2
sin D sin E
2
2
2
e
19.8
13.2 + 14.8 = c
=
sin 102° sin 37°
174.24 + 219.04 = c 2
e
19.8
=
393.28 = c 2
0.978 0.602
c = 19.8
19.8
× 0.602 = e
BE is 19.8 ft long.
0.978
e = 12.2
BD is 12.2 ft long.
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Step 3:
Using ∆BCD and the Law
of Sines, solve for ∠C .
d
c
=
sin D sin C
15.6
12.2
=
sin 72° sin C
15.6
12.2
=
0.951 sin C
15.6 sin C = 0.951 × 12.2
0.951 × 12.2
sin C =
15.6
sin C = 0.744
C = 48°
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C. D. Pilmer
Example 4
Nicholas and Matthew were training together for half-marathon. They had been running together
but in the last section, they decided to take different routes. Both turned off of Pilmer Boulevard.
Nicholas turned right onto Ormon Avenue and proceeded to the finish line. Matthew turned left
onto Jeffery Lane and proceeded to the finish line. The direct route along Pilmer Boulevard to
the finish line is 2.79 km. Which runner traveled further and by how much?
Ormon Avenue
(Nicholas’ Route)
finish line
38o
32o
Pilmer
Boulevard
o
29
Jeffery Lane
(Mathew’s Route)
Answer:
Nicholas’ Route
Since there are 180o in a triangle, we know that
the missing angle is 110o.
We need to find the two missing sides
(a and b) and add those values. We will use the
Law of Sines twice.
a
110o
38o
b
32o
c = 2.79 km
a
c
=
sin A sin C
a
2.79
=
sin 32° sin 110°
a
2.79
=
0.530 0.940
2.79
a = 0.530 ×
0.940
a = 1.57
b
c
=
sin B sin C
b
2.79
=
sin 38° sin 110°
b
2.79
=
0.616 0.940
2.79
b = 0.616 ×
0.940
b = 1.83
1.57 + 1.83 = 3.40
Nicholas’ route is 3.40 km long.
Matthew’s Route
We are dealing with a right angle triangle so we can rely on our three trigonometric ratios
(sine, cosine and tangent), rather than the Law of Sines.
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We need to find the two missing sides (e and f)
and add them together.
c = 2.79 km
29o
adj
hyp
opp
hyp
e
sin 29° =
2.79
e
0.485 =
2.79
0.485 × 2.79 = e
e = 1.35
cos θ =
1.35 + 2.44 = 3.79
Matthew’s route is 3.79 km long.
Matthew travels 0.39 km further.
sin θ =
e
f
f
cos 29° =
2.79
f
0.875 =
2.79
0.875 × 2.79 = f
f = 2.44
Note:
In many trigonometric word problems we will
encounter the terms angle of elevation and
angle of depression. These two terms are
used to describe the number of degrees one is
above or below the horizontal.
Angle of Elevation
Angle of Depression
horizontal
Questions:
1. (a) Solve for a.
(b) Solve for p.
R
B
26.8 cm
o
81
p
o
a
37
P
o
42
C
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7.89 m
40o
Q
A
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(c) Solve for ∠B .
(d) Solve for ∠G .
B
A
57.8 km
H
G
28o
4.82 m
21.8 km
73o
9.64 m
C
F
(f) Solve for ∠W .
(e) Solve for t.
V
R
104o
19.8 m
8.43 cm
t
o
T
68
32o
W
S
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U
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(h) Solve for ∠M .
(g) Solve for f.
F
43.7 m
7.64 km
106
M
L
o
G
32.8 m
49o
f
110o
N
H
2. Solve for ∠E and EH.
E
9.81 cm
F
8.74 cm
o
50
H
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32o
42o
G
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3. Solve for ∠N .
L
M
46o
52o
18.6 m
23.4 m
69o
120o
K
J
N
4. Solve for BC.
C
10.4 cm
D
57o
63o
E
B
112o
9.2 cm
35o
F
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5. A parallelogram has two sides (not adjacent sides) that are
14.8 cm and two interior angles that measure 68o. The
shorter diagonal is 23.7 cm. Determine the length of the
other two sides of the parallelogram. (Hint: This is a
multiple step problem.)
6. Three towns (Alton, Bridgeway and Colinville) are
located such that Bridgeway is 25.7 km from Alton,
and Colinville is 33.8 km from Alton. When one is
located at Bridgeway, the angle formed when viewing
between Alton and Colinville is 105o. How far is
Bridgeway from Colinville?
Bridgeway
Colinville
105o
Alton
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7. Two helicopters hovering at the same altitude spot a life raft bobbing in the ocean. The
helicopters are 3500 m apart. The angle of depression from the one helicopter to the raft is
28o. The angle of depression from the other helicopter to the raft is 36o. How far is each
helicopter away from the raft?
8. A radio tower is on level ground and supported by
two wires, as shown on the diagram. One wire is 16.7
m long and makes an angle of elevation of 48o. The
other wire is 13.4 m long. What is the angle of
elevation for this wire?
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Law of Cosines
The Law of Sines is a powerful tool but it has some limitations.
Case 1:
For example if you are given three sides of a triangle and
asked to find one of the interior angles (see diagram), the Law
of Sines does not allow you to solve this type of question.
Case 2:
Similarly the Law of Sines does not work when we are given
two sides that enclose a given interior angle and asked to find
the missing side (see diagram).
8.9 m
6.9 m
?
11.9 m
8.5 m
?
32o
12.3 m
These two cases show the need of another law; the Law of Cosines.
Deriving the Law of Cosines
We are only going to look at half of the derivation. We will show the derivation for acute
triangles (i.e. triangles where no interior angles exceeds 90o), but omit the derivation for obtuse
triangles.
We have started by taking ∆ABC and dividing into two right angle triangles.
A
b
c
h
B
C
D
m
n
a
For ∆ACD
b2 = h2 + n2
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For ∆ABD
n
b
n = b cos C
cos C =
c2 = h2 + m2
c 2 = h 2 + (a − n )
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We will now take the third equation, square the binomial, rearrange the terms, and incorporate
the other two equations.
c 2 = h 2 + (a − n )
c2
c2
c2
c2
c2
= h2
= a2
= a2
= a2
= a2
2
+ a 2 − 2an + n 2
+ h 2 + n 2 − 2an
+ h 2 + n 2 − 2an
+ b 2 − 2an
+ b 2 − 2ab cos C
(
)
The Law of Cosines:
We squared the binomial.
We rearranged the terms.
We grouped two of the terms together.
We substituted the first equation, b 2 = h 2 + n 2 , into our equation.
We substituted the second equation, n = b cos C , into our
equation.
c 2 = a 2 + b 2 − 2ab cos C
It can also be written as:
a 2 = b 2 + c 2 − 2bc cos A
or
2
2
b = a + c 2 − 2ac cos B
Checking the Law of Cosines
Below you have a triangle. Using a ruler and a protractor, measure the three sides and the three
angles. Record the values in the chart. Also complete the three calculations in the tables below
and on the next page.
A
B
c2
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a=
∠A =
b=
∠B =
c=
∠C =
C
a 2 + b 2 − 2ab cos C
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a2
b 2 + c 2 − 2bc cos A
b2
a 2 + c 2 − 2ac cos B
What do you notice? Is this what you expected?
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Using the Law of Cosines
In the previous section we were introduced to the Law of Cosines.
a 2 = b 2 + c 2 − 2bc cos A
2
B
c
b = a + c − 2ac cos B
2
2
c 2 = a 2 + b 2 − 2ab cos C
a
A
C
b
(Most textbooks will write the
equation in the first form.)
Example 1
Determine x.
8.6 m
x
o
32
13.7 m
Answer:
a 2 = b 2 + c 2 − 2bc cos A
x 2 = 8.6 2 + 13.7 2 − 2(8.6 )(13.7 ) cos 32°
x 2 = 73.96 + 187.69 − 235.64(0.8480 )
x 2 = 261.65 − 199.82
x 2 = 61.83
x = 61.83
x = 7.9
Example 2
Determine ∠P .
R
4.3 cm
P
5.7 cm
4.6 cm
Q
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Answer:
a 2 = b 2 + c 2 − 2bc cos A
p 2 = q 2 + r 2 − 2qr cos P
5.7 2 = 4.3 2 + 4.6 2 − 2(4.3)(4.6 ) cos P
32.49 = 18.49 + 21.16 − 39.56 cos P
32.49 − 18.49 − 21.16 = −39.56 cos P
− 7.16 = −39.56 cos P
− 7.16 − 39.56 cos P
=
− 39.56
− 39.56
0.1810 = cos P
P = 80°
Do not attempt to add 39.56 to both sides of the
equation.
We will now have to use the COS-1 button on
the calculator.
Example 3
Determine ∠L .
J
63o
15.3 m
K
17.9 cm
14.8 m
L
18.3 m
26o
M
N
Answer:
This is a three step problem. In the first step we will work with ∆JKN and use the Law of
Cosines to find KN.
a 2 = b 2 + c 2 − 2bc cos A
j 2 = k 2 + n 2 − 2kn cos J
j 2 = 17.9 2 + 15.3 2 − 2(17.9)(15.3) cos 63°
j 2 = 320.41 + 234.09 − 547.74(0.4540 )
j 2 = 554.50 − 248.67
j 2 = 305.83
j = 17.5 m
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In the second step we will work with ∆KMN and use the sine ratio to find KM.
opp
hyp
n
sin 26° =
17.5
n = 17.5 × sin 26°
n = 7.7 m
sin θ =
In the third and final step we will work with ∆KLM and use the Law of Cosines to find ∠L .
a 2 = b 2 + c 2 − 2bc cos A
l 2 = k 2 + m 2 − 2km cos L
7.7 2 = 18.3 2 + 14.8 2 − 2(18.3)(14.8) cos L
59.29 = 334.89 + 219.04 − 541.68 cos L
59.29 − 334.89 − 219.04 = −541.68 cos L
− 494.64 = −541.68 cos L
− 494.64 − 541.68 cos L
=
− 541.68
− 541.68
0.9132 = cos L
L = 24°
Example 4
Determine ∠EHF .
E
21.72 m
F
9.34 m
11.35 m
34o
H
69o
G
Answer:
This is a two step problem. In the first step we will work with ∆FGH and use the Law of
Sines to find FH.
g
h
=
sin G sin H
g
9.34
=
sin 69° sin 34°
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g
9.34
=
0.9336 0.5592
9.34
g = 0.9336 ×
0.5592
g = 15.59 m
In the second and final step we will work with ∆EFH and use the Law of Cosines to find
∠EHF .
a 2 = b 2 + c 2 − 2bc cos A
h 2 = e 2 + f 2 − 2ef cos H
21.72 2 = 15.59 2 + 11.35 2 − 2(15.59 )(11.35) cos H
471.76 = 243.05 + 128.82 − 353.89 cos H
471.76 − 243.05 − 128.82 = −353.89 cos H
99.89 = −353.89 cos H
99.89
− 353.89 cos H
=
− 353.89
− 353.89
− 0.2823 = cos H
H = 106°
Example 5
The straight line distance between Lei’s golf ball and the hole is 110 metres. Unfortunately she
makes a bad shot that is 20o to the right of the straight line path and the ball only travels 80
metres. How far is the ball from the hole after this bad shot?
Answer:
Draw a diagram and include all the relevant numbers.
hole
110 m
d
20o
80 m
Use the Law of Cosines to find the distance, d, between the ball and the hole after the bad
shot.
a 2 = b 2 + c 2 − 2bc cos A
d 2 = 110 2 + 80 2 − 2(110 )(80 ) cos 20°
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d 2 = 12100 + 6400 − 17600(0.9397 )
d 2 = 1961.28
d = 44.3 m
The ball is 44.3 m from the hole.
Interesting Fact:
You may have noticed that the Law of Cosines, when written in the form
c 2 = a 2 + b 2 − 2ab cos C , it looks a lot like that the Pythagorean Theorem. As we learned
previously, the Pythagorean Theorem only applies to right angle triangles. However, what
happens when we apply the Law of Cosines, rather than the Pythagorean Theorem, to a right
angle triangle? Does everything still work out? Consider the following example.
Example
Solve for the missing side using two different methods.
A
5.8 m
C
Answer:
Method 1: Law of Cosines
c 2 = a 2 + b 2 − 2ab cos C
10.8 m
B
Method 2: Pythagorean Theorem
c2 = a2 + b2
c 2 = 10.8 2 + 5.8 2 − 2(10.8)(5.8) cos 90°
c 2 = 10.8 2 + 5.8 2
c 2 = 116.64 + 33.64 − 125.28(0)
(Note that the cosine of 90o is zero.)
c 2 = 116.64 + 33.64
c 2 = 116.64 + 33.64
c 2 = 150.28
c = 12.3 m
c 2 = 150.28
c = 12.3 m
Notice that the Law of Cosines, although not the most efficient method, still generated the
correct answer. Based on this example and the derivation for the Law of Cosines seen in the
previous section, we can see that there is a connection between the Law of Cosines and the
Pythagorean Theorem.
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Questions
1. (a) Determine b.
C
b
A
113 cm
125 cm
42o
B
(b) Determine t.
R
S
31.2 km
102o
50.7 km
T
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(c) Determine ∠C .
B
141 km
A
314 km
230 km
C
(d) Determine ∠F .
F
21.6 cm
18.7 cm
E
26.5 cm
G
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(e) Determine p.
P
4.67 m
63o
Q
8.56 m
R
(f) Determine ∠M .
N
179 ft
127 ft
L
83 ft
M
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2. Solve for GH.
F
6.4 m
76o
E
D
o
48
10.9 m
72o
6.8 m
G
H
3. Determine ∠U .
S
30o
12.8 cm
R
110o
o
37
T
10.2 cm
U
V
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4. Determine ∠S .
P
57o
7.2 km
Q
18.3 km
T
42o
R
35o
12.4 km
S
5. A water molecule is comprised of one
oxygen atom and two hydrogen atoms.
The bonds between these atoms form a
triangle. The distance between these
atoms for water in its liquid form are
shown on the diagram. The units of
measure are picometres (pm).
Determine all the angles in our
molecular triangle.
Oxygen
96 pm
Hydrogen
96 pm
151.8 pm
Hydrogen
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6. A triangular lot sits at the corner of two streets
that intersect at an angle of 65o. One side of
the lot along one of the streets measures 45
metres. The other side along the other street
measures 48 metres. How long is the third
side of the lot?
lot
7. Anne and Dave are playing shuffleboard. Dave released the disk and it traveled 1.35 metres.
Anne released her disk from the same point and it traveled 1.17 metres. If the disks ended up
0.29 metres apart, what is the angle between the two paths of the two disks?
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Important Note:
In any Adult Learning Program course, time is a critical factor for many of our learners. Many
of our learners only have one year to complete their ALP courses before heading off to postsecondary studies. For this reason, the curriculum developers had to carefully consider the
concepts to be covered and the extent to which each of those concepts is explored. In the
sections on the Law of Cosines and the Law of Sines, the curriculum developers have chosen not
to examine the ambiguous case. This occurs when one is given a triangle where one acute angle
is supplied and two sides that do not enclose that angle are also supplied. This will makes more
sense when we consider an example.
Example
For ∆ABC , ∠A = 50° , b = 4.6 cm, and a = 3.7 cm.
(a) Find c.
(b) Find ∠B .
We are not going to solve this problem, but we will show how the supplied information can
be interpreted in two perfectly acceptable manners and thus result in two sets of acceptable
answers. If we take the one angle and the two sides and attempt to draw a triangle, we end up
with two valid interpretations.
Interpretation #1
b = 4.6 cm
Interpretation #2
b = 4.6 cm
a = 3.7 cm
50o
a = 3.7 cm
50o
A
A
If we used the Law of Cosines to find c, we would end up with the answers c = 4.08 cm
(Interpretation #1) and c = 1.83 cm (Interpretation #2). To accomplish this, we would have
to know how to solve quadratic equations, a topic that is covered later in this course.
If we used the Law of Sines to find ∠B , we would end up with the answers ∠B = 72 o
(Interpretation #1) and ∠B = 108o (Interpretation #2). To accomplish this, we would have to
understand the unit circle, a topic not covered in this course.
Although we will not be examining the ambiguous case in this course, it was still important
to mention in case we encounter it in a higher level math course.
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Putting It Together
1. Determine the indicated angle or side. Be prepared to use a variety of trigonometric tools
(i.e. Pythagorean Theorem, sine ratio, cosine ratio, tangent ratio, Law of Sines, or Law of
Cosines).
(b) Determine t.
(a) Determine ∠P .
R 4.73 m
T
Q
4.7 cm
1.2 cm
3.41 m
U
S
P
(c) Determine ∠N
L
(d) Determine f.
F
11.3 cm
35o
M
4.37 km
7.4 cm
62o
N
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(f) Determine ∠D .
E
(e) Determine j.
L
9.8 cm
12.7 m
49o
J
D
15.2 cm
K
(g) Determine ∠Q
P
(h) Determine h.
25.3 km
F
124 mm
H
o
107
Q
14.2 km
R
F
17.6 m
42o
27.8 km
G
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2. Determine ∠EDA .
E
D
68o
C
47o
15.3 m
12.7 m
56o
A
18.2 m
B
3. Determine ML .
M
L
63o
12.7 m
57o
N
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10.6 m
41o
13.9 m
K
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4. The face of a cliff rises vertically 46 m from the ocean’s surface. A boat offshore sights to
the top of the cliff and finds that the angle of elevation is 19o. How far is the boat from the
base of the cliff?
5. A marathon swimmer is swimming around three islands. She travels 2.6 km from Island A
to Island B. She then turns 55o and travels 2.9 km to island C. How long is the last leg of the
swim from Island C to Island A?
6. A telephone pole casts a shadow 8.5 m long when the angle of elevation of the sun is 52o.
(a) If the pole is vertical, determine the length of the exposed portion of the pole.
(b) If the pole is leaning away from the sun at an angle of 82 o from the horizontal, determine
the length of the exposed portion of the pole.
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7. A cable is used to support a vertical tower. The cable is attached to the top of the 42.0 m
tower and anchored to the ground 15.1 m from the base of the tower. How long is the cable?
8. In a last ditch effort to tie up a hockey game, the visiting
team has pulled their goalie. Unfortunately with only a few
seconds left in the game, one of the players for the home
team gets the puck and has a clear shot on the empty net.
The posts on a hockey net are 2.0 metres apart. From the
position that the home player shoots the puck, he is 8.2
metres from one post and 6.5 metres from the other post.
Within what angle, θ , must the shot be made to ensure a
goal?
θ
Player’s
Position
9. A hot air balloon is stationary at an altitude of 380
metres. The occupant of the balloon is looking
down at two objects on the ground. The angle of
depression to the first object is 69o. The angle of
depression to the second object is 46o. How far
apart are the two objects on the ground?
First
Object
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10. A golf hole has a dog leg around a
pond as shown in the diagram.
Suppose a golfer did not want to use
the dog leg and instead wanted to
shoot over the pond.
(a) How far would the golfer have to
drive the ball to make it to the
hole?
(b) How many degrees to the left of
the proposed path would the golfer
have to shoot if he/she intends to
put the ball on the green in one
shot rather than two shots?
65.0 m
115o
89.0 m
pond
tee
11. Jacob is hovering in a helicopter and looking down at
two friends’ homes. The angle of depression to the
first home is 72o. The angle of depression to the second
home is 49o. If the homes are 420 metres apart, what is
the distance between the helicopter and the first home?
(Hint: The angle of depression from the helicopter to
the first home is equal to the angle of elevation from the
first home to the helicopter. The relationship exists
between the angle of elevations and depressions
associated with the second home and the helicopter.)
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Post-Unit Reflections
What is the most valuable or important
thing you learned in this unit?
What part did you find most interesting or
enjoyable?
What was the most challenging part, and
how did you respond to this challenge?
How did you feel about this math topic
when you started this unit?
How do you feel about this math topic
now?
Of the skills you used in this unit, which
is your strongest skill?
What skill(s) do you feel you need to
improve, and how will you improve them?
How does what you learned in this unit fit
with your personal goals?
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Which Formula Should I Use?
In this unit we have worked with six different formulas.
Pythagorean Theorem:
a2 + b2 = c2
Three Trigonometric Ratios:
opp
adj
sin θ =
cos θ =
hyp
hyp
Law of Sines:
a
b
c
=
=
sin A sin B sin C
Law of Cosines:
a 2 = b 2 + c 2 − 2bc cos A
tan θ =
opp
adj
Deciding which formula to use comes down to understanding the seven types of triangle
questions that can be asked. There are only three types of questions that can be asked about
right-angle triangles, and only four types of questions that can be asked about triangles that are
not right-angled.
It is important to note that when we are given a right-angle triangle problem, you are typically
given two pieces of information (other than the right angle) and asked to find a third.
• We use the Pythagorean Theorem when we are given two sides of a right-angle triangle
and asked to find the third side.
• We use one of our three trigonometric ratios when we are given two sides of a right-angle
triangle and asked to find an angle.
• We use one of our three trigonometric ratios when we are given one side and one angle of
a right-angle triangle, and asked to find an unknown side.
It is important to note that when we are dealing with questions involving triangles that are not
right-angled, we are typically given three pieces of information, and asked to find a fourth.
• We use the Law of Sines when we are given two sides and one corresponding angle, and
asked to find the other corresponding angle.
• We use the Law of Sines when we are given two angles and one corresponding side, and
asked to find the other corresponding side.
• We use the Law of Cosines when we are given three sides, and asked to find an angle.
• We use the Law of Cosines when we are given one angle and two sides, and asked to find
the corresponding side.
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Answers
Do not be alarmed if your answers throughout this unit are off by a few tenths or hundredths
compared to the answers supplied in this resource. These small variations are usually due to the
rounding procedures you decide to use. Your instructor is more concerned with you using the
correct procedure and understanding the concepts.
The Pythagorean Theorem (pages 1 to 6)
1. (a) 14.0 cm
(c) 14.8 cm
(b) 13.5 m
(d) 24.2 ft
2. It is right-angled because 8 2 + 15 2 = 17 2
3. (a) 10.6
(c) 10.8
(b) 9.8
4. 6.8 m
5. 11.3 m
6. 1.9 km
Trigonometric Ratios (pages 7 to 10)
Investigation
Part 1
Triangle
Opposite
(20o)
Adjacent
Hypotenuse
Opposite
Hypotenuse
Adjacent
Hypotenuse
Opposite
Adjacent
∆ABC
19 mm
57 mm
60 mm
0.32
0.95
0.33
∆ADE
26 mm
74 mm
79 mm
0.33
0.94
0.35
∆AFG
32 mm
92 mm
97 mm
0.33
0.95
0.35
Opposite
Adjacent
Hypotenuse
Opposite
Hypotenuse
Adjacent
Hypotenuse
Opposite
Adjacent
∆ABC
26 mm
47 mm
53 mm
0.49
0.89
0.55
∆ADE
36 mm
65 mm
74 mm
0.49
0.88
0.55
∆AFG
45 mm
82 mm
93 mm
0.48
0.88
0.55
Part 2
Triangle
(30o)
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Part 3
Triangle
(40o)
Opposite
Adjacent
Hypotenuse
Opposite
Hypotenuse
Adjacent
Hypotenuse
Opposite
Adjacent
∆ABC
35 mm
44 mm
56 mm
0.63
0.79
0.80
∆ADE
48 mm
60 mm
77 mm
0.62
0.78
0.80
∆AFG
61 mm
76 mm
97 mm
0.63
0.78
0.80
Multiple Choice Questions
1. (d)
2. (c)
3. (b)
4. (a)
5. (c)
Conclusions
Sine Ratio:
sin θ =
opposite
hypotenuse
Cosine Ratio:
cos θ =
adjacent
hypotenuse
Tangent Ratio:
tan θ =
opposite
adjacent
Using Our Three Trigonometric Ratios, Part 1 (pages 11 to 16)
x
, answer: 9.6
16.7
9.2
, answer: 22o
tan θ =
22.7
x
, answer: 29.4
cos 21° =
31.5
54 o
159.5
10.4
1. (a) sin 35° =
(c)
(e)
(g)
(i)
(k)
9.8
, answer: 19.2
x
5.7
, answer: 26 o
sin θ =
12.9
10.3
, answer: 65 o
cos θ =
24.7
0.89
47 o
68 o
(b) tan 27° =
(d)
(f)
(h)
(j)
(l)
Using Our Three Trigonometric Ratios, Part 2 (pages 17 to 21)
1. First Step: 15.9 m, Final Step: 31.2 m
2. First Step: 98.3 cm, Final Step: 24o
3. First Step: 53.1 m, Second Step: 25.6 m, Final Step: 41.6 m
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4. First Step: 17.3 km, Second Step: x = 25.6 km, Third Step: 30.9 km, Final Step: θ = 16o
5. 81o
6. 10.4 + 12.3 = 22.7 m
7. 105 m
8. 138.8 m
9. 95.4 miles
10. 60.4 – 55.1 = 5.3 m
Law of Sines (pages 22 to 24)
a = 5.95 cm
∠A = 62°
b = 6.42 cm
∠B = 74°
c = 4.66 cm
∠C = 44°
a
= 6.7
sin A
b
= 6.7
sin B
c
= 6.7
sin C
We noticed that all three calculations were equal to 6.7 (or approximately 6.7 depending on the
accuracy of your measurements). We expected that these three calculations would be equal
because they represented the three components of the Law of Sines.
Using the Law of Sines (pages 25 to 33)
1. (a)
(c)
(e)
(g)
5.35 m
70o
34.6 m
9.73 km
(b)
(d)
(f)
(h)
25.1 cm
21 o
37 o
45o
2. First Step: 11.04 cm, Second Step: ∠E = 60° , Final Step: EH = 12.03 cm
3. First Step: ∠L = 65° , Second Step: KM = 23.4 m, Third Step: JM = 14.4 m,
Final Step: ∠N = 32°
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4. First Step: CE = 9.8 cm, Second Step: ∠EBF = 33° , Third Step: BE = 9.7 cm,
Final Step: BC = 13.8 cm
5. First Step: 35 o, Second Step: 77 o, Final Step: 24.9 cm
6. First Step: 47 o, Second Step: 28 o, Final Step: 16.4 km
7. 2289 m and 1828 m
8. 68 o
Law of Cosines (pages 34 to 36)
a = 5.95 cm
∠A = 62°
b = 6.42 cm
∠B = 74°
c = 4.66 cm
∠C = 44°
c2
a 2 + b 2 − 2ab cos C
4.66 2
21.7
5.95 2 + 6.42 2 − 2(5.95)(6.42 ) cos 44°
35.4025 + 41.2164 − 76.398(0.7193)
76.6189 − 54.9531
21.7
a2
b 2 + c 2 − 2bc cos A
5.95 2
35.4
6.42 2 + 4.66 2 − 2(6.42 )(4.66 ) cos 62°
41.2164 + 21.7156 − 59.8344(0.4695)
62.9320 − 28.0923
34.8
b2
a 2 + c 2 − 2ac cos B
6.42 2
41.2
5.95 2 + 4.66 2 − 2(5.95)(4.66 ) cos 74°
35.4025 + 21.7156 − 55.454(0.2756 )
57.1181 − 15.2831
41.8
We noticed that column1 and column 2 of each table are approximately equal. If we could
accurately measure each side and angle, then the results would have been equal, rather than
approximately equal. We expected that both columns would be equal because we working out
both sides of the equation for the Law of Cosines.
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Using the Law of Cosines (pages 37 to 48)
1. (a) 86 cm
(c) 24o
(e) 7.67 m
(b) 64.8 km
(d) 82 o
(f) 115 o
2. First Step: EH = 5.3 m, Second Step: EG = 11.2 m, Final Step: GH = 9.9 m
3. First Step: RT = 6.8 cm, Second Step: TV = 5.1 cm, Final Step: ∠U = 30 o
4. First Step: QT = 15.6 km, Second Step: RT = 11.6 km, Final Step: ∠S = 32 o
5. 104 o, 38 o, and 38 o
6. 50 m
7. 10 o
Putting It Together (pages 49 to 54)
1. (a)
(c)
(e)
(g)
54o
61o
11.5 cm
31o
(b)
(d)
(f)
(h)
4.5 cm
2.05 km
46o
177 mm
2. First Step: BD = 17.4 m, Second Step: AD = 16.7 m, Final Step: ∠EDA = 58°
3. First Step: JM = 11.3 m, Second Step: JL = 10.2 m, Final Step: ML = 15.2 m
4. 134 m
5. 2.6 km
6. (a) 10.9 m
(b) 9.3 m
7. 44.6 m
8. 8o
9. 221 m
10. (a) 130.5 m
(b) 27o
11. 811 m
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