Download Trigonometry Unit

Trigonometry Unit
(Level IV Graduate Math)
Draft
(NSSAL)
C. David Pilmer
©2010
(Last Updated: Dec 2011)
This resource is the intellectual property of the Adult Education Division of the Nova Scotia
Department of Labour and Advanced Education.
The following are permitted to use and reproduce this resource for classroom purposes.
• Nova Scotia instructors delivering the Nova Scotia Adult Learning Program
• Canadian public school teachers delivering public school curriculum
• Canadian nonprofit tuition-free adult basic education programs
The following are not permitted to use or reproduce this resource without the written
authorization of the Adult Education Division of the Nova Scotia Department of Labour and
Advanced Education.
• Upgrading programs at post-secondary institutions
• Core programs at post-secondary institutions
• Public or private schools outside of Canada
• Basic adult education programs outside of Canada
Individuals, not including teachers or instructors, are permitted to use this resource for their own
learning. They are not permitted to make multiple copies of the resource for distribution. Nor
are they permitted to use this resource under the direction of a teacher or instructor at a learning
institution.
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 …………………………………………………………………………….
ii
ii
iii
iv
Prerequisite Knowledge ……………………………………………………………………..
Investigating Similar Triangles ………………………………………………………………
Using Similar Triangles ……………………………………………………………………...
Similar Triangle Application Problems ……………………………………………………...
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
5
7
16
20
26
30
36
42
46
56
59
72
Post Unit Reflection …………………………………………………………………………. 79
Answers ……………………………………………………………………………………… 80
NSSAL
©2010
i
Draft
C. D. Pilmer
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 only focus on the relationships between the sides and angles of triangles.
Initially we will work with similar triangles. Then 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. Finally 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 right-angle triangles).
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:
NSSAL
©2010
__________________________
ii
Draft
C. D. Pilmer
The Big Picture
The following flow chart shows the five required units and the four optional units (choose two of
the four) in Level IV Graduate Math. These have been presented in a suggested order.
Math in the Real World Unit (Required)
• Fractions, decimals, percents, ratios, proportions, and
signed numbers in real world applications
• Career Exploration and Math
Solving Equations Unit (Required)
• Solve and check equations of the form Ax + B = Cx + D ,
A = Bx 2 + C , and A = Bx 3 + C .
Consumer Finance Unit (Required)
• Simple Interest and Compound Interest
• TVM Solver (Loans and Investments)
• Credit and Credit Scores
Graphs and Functions Unit (Required)
• Understanding Graphs
• Linear Functions and Line of Best Fit
Measurement Unit (Required)
• Imperial and Metric Measures
• Precision and Accuracy
• Perimeter, Area and Volume
Choose two of the four.
Linear
Functions and
Linear Systems
Unit
Statistics Unit
Trigonometry
Unit
ALP Approved
Projects
(Complete 2 of
the 5 projects.)
Note:
You are not permitted to complete four ALP Approved Projects and thus avoid selecting from
the Linear Functions and Linear Systems Unit, Trigonometry Unit, or Statistics Unit.
NSSAL
©2010
iii
Draft
C. D. Pilmer
Course Timelines
Graduate 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.
Unit Name
Minimum
Completion Time
in Hours
24
20
18
28
24
20
20
Total: 154 hours
Math in the Real World Unit
Solving Equations Unit
Consumer Finance Unit
Graphs and Functions Unit
Measurement Unit
Selected Unit #1
Selected Unit #2
Maximum
Completion Time
in Hours
36
28
24
34
30
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.
NSSAL
©2010
iv
Draft
C. D. Pilmer
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°
50o
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.
B
C
b
A
Complementary Angles
Two angles are complementary if they add up to 90o (i.e. form a
right angle). In the diagram ∠ABC and ∠CBD are
complementary angles.
a
c
B
A
C
B
Supplementary Angles
Two angles are supplementary if they add up to 180o (i.e. form a
straight line). In the diagram ∠DEF and ∠FEG are
supplementary angles.
D
F
D
Opposite Angles
When two lines intersect, four angles are formed. The angles that
are directly opposite to each other are called opposite angles.
These opposite angles are congruent (i.e. equal in measure). In the
diagram ∠ABE = ∠CBD and ∠ABC = ∠DBE .
NSSAL
©2010
1
E
G
A
C
B
E
D
Draft
C. D. Pilmer
Right Angle Triangles
A right angle triangle has one angle measuring 90o (i.e. a right angle).
In the diagram, ∆ABC is a right angle triangle because ∠ABC
measures 90o.
A
B
Acute Triangles
An acute triangle is a triangle where all three angles are less than
90o. In the diagram, ∆FGH is an acute triangle because
∠FGH = 70° , ∠HFG = 47° , and ∠GHF = 63° (All the angles
are less than 90o).
C
F
G
H
Obtuse Triangles
An obtuse triangle is a triangle where one of the three angles is
greater than 90o. In the diagram, ∆PQR is an obtuse triangle
because the ∠QPR = 107° (It exceeds 90o.)
R
Q
P
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°
NSSAL
©2010
2
C
A
B
H
G
E
F
Draft
C. D. Pilmer
Parallel Lines and Transversal Lines
Lines are parallel if they are always the same distance apart
and never meet. We use arrow symbols to indicate when
lines are parallel. When parallel lines are crossed by
another line (called a transversal line), angles are formed
and special relationships exist among these angles.
F
E
D
I
H
Angle Relationships:
G
K
J
1. Corresponding angles are equal.
∠FED = ∠EIH , ∠DEI = ∠HIJ , ∠FEG = ∠EIK , and
∠GEI = ∠KIJ
2. Alternate interior angles are equal.
∠DEI = ∠EIK and ∠GEI = ∠EIH
3. Alternate exterior angles are equal.
∠FED = ∠JIK and ∠FEG = ∠JIH
4. Consecutive interior angles are supplementary.
∠DEI + ∠EIH = 180°
∠GEI + ∠EIK = 180°
Prerequisite Practice (Optional)
Find the identified angles. (Diagrams are not to scale.)
1.
2.
E
D
M
N
40o
A
B
75
Q
P
o
95
O
C
40o
∠DBC =
∠ONP =
∠EBD =
∠OPN =
∠ABC =
∠NPQ =
∠BDC =
∠QPR =
NSSAL
©2010
o
3
R
Draft
C. D. Pilmer
3.
4.
B
C
G
H
H
40o
115o
D 95o
o
70
o
95
E
F
I
L
55o
N
J
G
∠GHF =
∠ILM =
∠HFC =
∠MLN =
∠CFE =
∠KLN =
∠DCF =
∠KNL =
∠BCD =
∠HIL =
∠EFG =
∠JIL =
NSSAL
©2010
K
4
M
Draft
C. D. Pilmer
Investigating Similar Triangles
Below we have been provided with two similar triangles. At this point we do not know the
formal definition of a similar triangle. We going to examine these two triangles and try to see
how they are related and why they are called similar triangles.
C
A
B
F
D
E
Step 1: Using a protractor, measure the three interior angles of each triangle.
Triangle #1
∠CAB =
∠ABC =
∠BCA =
Triangle #2
∠FDE =
∠DEF =
∠EFD =
What do you notice?
NSSAL
©2010
5
Draft
C. D. Pilmer
Step 2: Using a ruler, measure the length of each side of each triangle to the nearest millimeter.
Triangle #1
AB =
BC =
AC =
Triangle #2
DE =
EF =
DF =
Step 3: We are going to use the measurements from Step 2 to work out three ratios.
First Ratio: Take the length of AB and divide it by the length of DE.
AB
=
DE
=
Second Ratio: Take the length of BC and divide it by the length of EF.
BC
=
EF
=
Third Ratio: Take the length of AC and divide it by the length of DF.
AC
=
DF
=
What do you notice?
Step 4: Based on what you have done, how can you tell if two triangles are similar?
NSSAL
©2010
6
Draft
C. D. Pilmer
Using Similar Triangles
Similar Triangle Definition
In any pair of similar triangles the corresponding angles are equal, and the ratio of the
corresponding sides are equal. For our diagram below, ∆ABC and ∆DEF are similar triangles
because
∠CAB = ∠FDE
∠ABC = ∠DEF
∠BCA = ∠EFD
and
AB BC AC
=
=
DE EF DF
F
C
D
A
B
E
Identifying Similar Triangles
Sometimes it is little more difficult to identify similar triangles. We often have to recognize
relationships between angles, before we can conclude that we are dealing with similar triangles.
Consider the following diagrams.
Diagram 1
A
∆ACE and ∆BCD are similar triangles.
B
E
NSSAL
©2010
D
C
Why?
1. ∠ACE = ∠BCD because they are the same
angle.
2. ∠CEA = ∠CDB because they are both labeled
as right angles.
3. ∠EAC = ∠DBC because there are only 180o
in a triangle and if the two triangles have two
sets of equal angles, then the third set of angles
must be equal.
4. Since all the corresponding angles are equal,
then we are dealing with similar triangles.
7
Draft
C. D. Pilmer
Diagram 2
S
R
Q
T
P
∆PQR and ∆STR are similar triangles.
Why?
1. ∠PRQ = ∠SRT because they are opposite
angles.
2. ∠RQP = ∠RTS because they are both labeled
as right angles.
3. ∠QPR = ∠TSR because there are only 180o in
a triangle and if the two triangles have two sets
of equal angles, then the third set of angles
must be equal.
4. Since all the corresponding angles are equal,
then we are dealing with similar triangles.
Diagram 3
J
I
∆FGJ and ∆FHI are similar triangles.
F
G
H
Why?
1. ∠GFJ = ∠HFI because they are the same
angle.
2. ∠FGJ = ∠FHI because when dealing with
parallel lines cut by a transversal line,
corresponding angles are equal.
3. ∠FJG = ∠FIH because when dealing with
parallel lines cut by a transversal line,
corresponding angles are equal.
4. Since all the corresponding angles are equal,
then we are dealing with similar triangles.
Example 1
If ∆ABC and ∆DEF are similar triangles, find AB given the information supplied in the
diagram.
D
A
B
NSSAL
©2010
31.7 cm
22.3 cm
C
E
43.5 cm
8
F
Draft
C. D. Pilmer
Answer:
Since we are dealing with similar triangles, we know the following.
AB BC AC
=
=
DE EF DF
For the information provided, we are only going to use the following.
AB BC
=
DE EF
Now we will substitute the known values into our equation and solve for the unknown value
(i.e. the length of side AB). Rather than calling our unknown side AB, we can call it side
c (note we are using a lowercase c) because it is opposite (or across from) ∠C .
AB BC
=
DE EF
c
22.3
=
31.7 43.5
43.5c = 22.3 × 31.7
43.5c = 706.91
706.91
c=
43.5
c = 16.3 cm
Procedure:
1. Find the cross products.
2. Set the cross products equal to each other.
3. Divide the number not multiplied by variable by
the number multiplied by the variable (in this case
the variable is c).
Example 2
If ∆JKL and ∆MNO are similar triangles, find NO given the information supplied in the
diagram.
O
L
18.9 m
12.3 m
20.1 m
J
K
M
N
Answer:
JL
KL
=
MO NO
12.3 20.1
=
18.9
m
12.3m = 20.1 × 18.9
12.3m = 379.89
NSSAL
©2010
Rather than calling our unknown side NO, we can call
it side m because it is opposite (or across from) ∠M .
Often people will use the variable x to represent the
unknown side. Feel free to do that if you find it easier.
9
Draft
C. D. Pilmer
379.89
12.3
m = 30.9 m
m=
Example 3
In the diagram below, QS equals 62 mm, QP equals 21 mm, and RT equals 17 mm. Find RS.
P
T
S
57o
57o
R
Q
Answer:
We suspect that we are dealing with similar triangles. We have to start by proving this
suspicion.
•
•
•
∠PQS = ∠TRS because they are both 57o.
∠PSQ = ∠TSR because they are the same angle.
∠SPQ = ∠STR because there are 180o in any triangle.
Since all corresponding angles are equal, we now know that we are dealing with similar
triangles.
RS RT
=
QS QP
t
17
=
62 21
21t = 17 × 62
21t = 1054
1054
t=
21
t = 50.2 mm
NSSAL
©2010
10
Draft
C. D. Pilmer
Example 4
In the diagram below, EF equals 51.2 cm, FH equals 71.8 cm, and DE equals 168.3 cm.
Determine GH.
D
E
F
G
H
Answer:
We suspect that we are dealing with similar triangles. We have to start by proving this
suspicion.
•
•
•
∠EFD = ∠HFG because they are opposite angles.
∠DEH = ∠EHG because when dealing with parallel lines cut by a transversal line
(EH), alternate interior angles are equal.
∠EDG = ∠HGD because when dealing with parallel lines cut by a transversal line
(DG), alternate interior angles are equal.
Since all corresponding angles are equal, we now know that we are dealing with similar
triangles.
DE EF
=
GH HF
168.3 51.2
=
71.8
GH
51.2GH = 168.3 × 71.8
51.2GH = 12083.94
12083.94
GH =
51.2
GH = 236.0 cm
NSSAL
©2010
In this question we did not attempt to rename
side GH. The reason was that vertex F is
shared by both triangles so if we attempted to
name a side f, we would not know if we were
referring to side GH or side DE.
Often people will use the variable x to
represent the unknown side. Feel free to do
that if you find it easier.
11
Draft
C. D. Pilmer
Questions:
1. If these two triangles are similar, find PQ.
5.6 m
E
G
8.2 m
P
R
10.8 m
F
Q
2. If the following triangles are similar, determine the measure of CD.
S
C
B
5.2 km
R
7.8 km
7.3 km
D
T
NSSAL
©2010
12
Draft
C. D. Pilmer
3. If these two triangles are similar, find TV and TU.
C
T
8.1 km
U
19.4 km
E
8.8 km
14.1 km
V
D
4. If the following triangles are similar, determine the measures of FG and EF.
8.7 m
J
G
K
10.8 m
5.1 m
7.4 m
L
E
NSSAL
©2010
F
13
Draft
C. D. Pilmer
5. Find DE.
C
E
96 cm
97o
D
141 cm
97o
179 cm
F
B
6. Given that JK = 77 m, KN = 36 m, JN = 85 m, and JL = 116 m, find LM and JM.
K
L
J
N
M
NSSAL
©2010
14
Draft
C. D. Pilmer
7. If PN = 78 cm, PQ = 43 cm, NO = 48 cm, and OP = 86 cm, find QR and PR.
O
P
Q
N
R
8. If WV = 4.6 m, UW = 6.8 m, ST = 13.7 m, and UV = 6.5 m, find SV and TV.
S
T
U
W
V
NSSAL
©2010
15
Draft
C. D. Pilmer
Similar Triangle Application Questions
We will now look at real world problems that can be solved using similar triangles.
Example 1
George is standing in a flat field. A 1.6 m height pole located in that field casts a shadow that is
2.3 m long. At the same time, a tree in the same field casts a 11.7 m long shadow. What is the
height of the tree?
Answer:
We will start by drawing a diagram.
1.6 m
11.7 m
2.3 m
Since we are dealing with two vertical objects (the tree and pole) and shadows cast at the
same time of day, we can conclude that all corresponding angles are equal. If this is the case
then we are dealing with similar triangles.
h 11.7
=
1.6 2.3
2.3h = 11.7 × 1.6
2.3h = 18.72
18.72
2.3
h = 8.1 m
h=
NSSAL
©2010
The height of the tree is 8.1 m.
16
Draft
C. D. Pilmer
Example 2
A mirror is placed on the floor between a wall and a small laser. The beam of the laser is pointed
at a mirror. The beam is reflected off the mirror and strikes the top of a wall. The distance along
the floor between the mirror and the wall is 3.5 metres. The distance along the floor between the
mirror and the vertical stand holding the laser is 0.9 m. If the laser is 1.5 m off the floor, how tall
is the wall. Please note that the angle the laser beam strikes the mirror is equal to the angle the
laser beam reflects to the top of the wall.
Answer:
Start by drawing a diagram.
wall
laser
1.5 m
mirror
3.5 m
0.9 m
If we just look at the similar triangles, we end up with the following.
0.9 1.5
=
3.5
h
0.9h = 1.5 × 3.5
0.9h = 5.25
h
1.5 m
3.5 m
NSSAL
©2010
5.25
0.9
h = 5.8 m
h=
0.9 m
17
Draft
C. D. Pilmer
Questions:
1. A company, that manufactures extension ladders, recommends for safety reasons that the
base of the ladder should be 1 metre away from the wall for every 4 metres the ladder rises
vertically. If the ladder must reach the top of a 7 metre wall, how far should the base of the
ladder be away from the base of this wall?
2. Ryan wants to know the width of a river but he has no means to take that measurement
directly. He decides to make some measurements along the shoreline to accomplish this
task. The distance from A to C is 161 m. The distance from B to C is 63 m. The distance
from C to D is 37 m. Determine the width of the river.
River
A
B
C
D
NSSAL
©2010
18
Draft
C. D. Pilmer
3. At a certain time of day, the shadow cast by a flag pole is 10.42 m long. At the same time,
the shadow cast by 1.85 m high individual is 1.27 m. How tall is the flag pole?
4. A wheel chair ramp has been built for the new community centre. If someone travels 2
metres horizontally along the ramp, they rise 0.18 m. How much would one rise, if they
traveled 4.5 metres along the same ramp?
5. Montez is building a cottage and has to attach two
small pieces of wood to the gable end (see shaded
triangles in diagram). The triangular pieces that are
18 inches long must have the same pitch as his roof.
The roof’s pitch is 8:12 which means that for every 12
feet one travels horizontally, the roof goes up by 8
feet. How tall should the triangular piece of wood be?
NSSAL
©2010
19
Draft
C. D. Pilmer
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
NSSAL
©2010
20
3.9 m
Q
2.3 m
q
R
Draft
C. D. Pilmer
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.
NSSAL
©2010
21
Draft
C. D. Pilmer
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
NSSAL
©2010
16.7 m
22
Draft
C. D. Pilmer
(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
NSSAL
©2010
23
Draft
C. D. Pilmer
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
NSSAL
©2010
24
Draft
C. D. Pilmer
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?
NSSAL
©2010
25
Draft
C. D. Pilmer
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
NSSAL
©2010
20o
C
E
26
G
Draft
C. D. Pilmer
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?
NSSAL
©2010
27
Draft
C. D. Pilmer
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
NSSAL
©2010
28
Draft
C. D. Pilmer
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:
NSSAL
©2010
cos θ =
opposite
θ
adjacent
tan θ =
29
Draft
C. D. Pilmer
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
NSSAL
©2010
30
x
opp
8.6
hyp
25o
Draft
C. D. Pilmer
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
NSSAL
©2010
31
Draft
C. D. Pilmer
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°
NSSAL
©2010
8.3
opp
θ
11.9
adj
32
Draft
C. D. Pilmer
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
NSSAL
©2010
33
Draft
C. D. Pilmer
(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
NSSAL
©2010
34
Draft
C. D. Pilmer
(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
NSSAL
©2010
35
Draft
C. D. Pilmer
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.
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.
Understanding this is really important when attempting to handle a multiple step question.
Example 1
Solve for x.
B
C
o
x
22
31o
A
94 m
E
176 m
D
Answer:
In this question we have three right-angle triangles that are connected together.
Although the question asks us to solve for x, we cannot start by doing so because we are only
given one piece of information (other than the right-angle) for ∆ABE . We only know the
length of one side. Side AE is 94 m.
Can we use ∆BCE is solve for side BE , the side shared by ∆ABE and ∆BCE ?
Unfortunately not; we are only given one piece of information about ∆BCE . We know that
one of its angles measures 22o.
We only have one triangle left. For ∆CDE we are given two pieces of information; a side
and an angle. We could use this information to find side CE, the side shared by ∆CDE and
∆BCE .
Now things start to fall into place. Once we have solved for CE using ∆CDE , we can solve
for BE using ∆BCE , and finally solve for for AB (or x) using ∆ABE . So we have to do three
sets of calculations to complete this question.
NSSAL
©2010
36
Draft
C. D. Pilmer
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
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
176
0.857
CE = 205.4
BE = 77.0
CE =
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.
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.
NSSAL
©2010
37
Angle of Elevation
Angle of Depression
horizontal
Draft
C. D. Pilmer
Questions
1. Solve for x.
Hint:
Please note that this figure is comprised
of three right-angle triangles. You are
going to start by finding the hypotenuse
of the smallest right-angle triangle. Once
this is complete you can solve for x using
the largest right-angle triangle.
x
63o
7.2 m
2. Solve for θ .
θ
153 cm
NSSAL
©2010
50o
241 cm
38
Draft
C. D. Pilmer
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
NSSAL
©2010
8.5 km
39
Draft
C. D. Pilmer
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.
NSSAL
©2010
40
post
35o
150 m
Draft
C. D. Pilmer
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?
NSSAL
©2010
41
Draft
C. D. Pilmer
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.
NSSAL
©2010
42
Draft
C. D. Pilmer
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
NSSAL
©2010
This is the Law of Sines.
43
Draft
C. D. Pilmer
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?
NSSAL
©2010
44
Draft
C. D. Pilmer
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
NSSAL
©2010
45
Draft
C. D. Pilmer
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°
It is important to note that when we are dealing with a Law of Sine questions, we are 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.
This is important to know when we are dealing with multiple step questions.
Example 3
Determine ∠C .
14.8 ft
B
A
15.6 ft
C
13.2 ft
72o
37o
102o
D
E
Answer:
We cannot start by solving for ∠C because we only have two pieces of information for
∆BCD , rather than the required three pieces of information.
Although side BD is shared by two triangles, we don't have enough information about
∆BDE to solve for side BD.
∆ABE is a right-angle triangle. For these triangles we only need to be given two pieces of
information (other than the right-angle) in order to solve for a third. We could use the
Pythagorean Theorem to solve for side BE, the side shared by ∆ABE and ∆BDE .
Once we have BE, we can solve for BD and then solve for ∠C .
NSSAL
©2010
46
Draft
C. D. Pilmer
It takes three steps to solve this question.
Step 1:
Using ∆ABE and the
Pythagorean Theorem, solve
for BE.
a2 + b2 = c2
13.2 2 + 14.8 2 = c 2
174.24 + 219.04 = c 2
393.28 = c 2
c = 19.8
BE is 19.8 ft long.
Step 2:
Using ∆BDE and the Law of
Sines, solve for BD.
d
e
=
sin D sin E
e
19.8
=
sin 102° sin 37°
19.8
e
=
0.978 0.602
19.8
× 0.602 = e
0.978
e = 12.2
BD is 12.2 ft long.
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°
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
o
29
Pilmer
Boulevard
Jeffery Lane
(Mathew’s Route)
NSSAL
©2010
47
Draft
C. D. Pilmer
Answer:
Nicholas’ Route
Since there are 180o in a triangle, we know that
the missing angle is 110o.
a
We need to find the two missing sides
(a and b) and add those values. We will use the
Law of Sines twice.
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.
We need to find the two missing sides (e and f)
c = 2.79 km
and add them together.
29o
adj
opp
cos θ =
sin θ =
hyp
hyp
e
f
f
e
cos 29° =
sin 29° =
2.79
2.79
f
e
0.875 =
0.485 =
2.79
2.79
0.875 × 2.79 = f
0.485 × 2.79 = e
e = 1.35
1.35 + 2.44 = 3.79
NSSAL
©2010
f = 2.44
Matthew’s route is 3.79 km long.
Matthew travels 0.39 km further.
48
Draft
C. D. Pilmer
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
42o
C
NSSAL
©2010
7.89 m
40o
Q
A
49
Draft
C. D. Pilmer
(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
NSSAL
©2010
50
U
13.45 cm
Draft
C. D. Pilmer
(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
(Hint: Start by solving for FH.)
9.81 cm
F
8.74 cm
o
50
H
NSSAL
©2010
32o
42o
G
51
Draft
C. D. Pilmer
3. Solve for ∠N .
L
M
46o
52o
18.6 m
23.4 m
69o
120o
K
J
N
NSSAL
©2010
52
Draft
C. D. Pilmer
4. Solve for BC.
C
10.4 cm
D
57o
63o
E
B
o
112
9.2 cm
35o
F
NSSAL
©2010
53
Draft
C. D. Pilmer
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
NSSAL
©2010
54
Draft
C. D. Pilmer
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?
NSSAL
©2010
55
Draft
C. D. Pilmer
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
NSSAL
©2010
For ∆ABD
n
b
n = b cos C
cos C =
c2 = h2 + m2
c 2 = h 2 + (a − n )
56
2
Draft
C. D. Pilmer
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
NSSAL
©2010
a=
∠A =
b=
∠B =
c=
∠C =
C
a 2 + b 2 − 2ab cos C
57
Draft
C. D. Pilmer
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?
NSSAL
©2010
58
Draft
C. D. Pilmer
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
NSSAL
©2010
59
Draft
C. D. Pilmer
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.
It is important to note that when we are dealing with a Law of Cosine questions, we are given
three pieces of information, and asked to find a fourth.
• 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.
This is important to know when we are dealing with multiple step questions.
Example 3
Determine ∠L .
J
63o
15.3 m
17.9 cm
K
14.8 m
L
18.3 m
26o
N
M
Answer:
We cannot start by solving for ∠L because we only have two pieces of information for
∆KLM , rather than the required three pieces of information.
∆KMN is a right-angle triangle which shares side KM with ∆KLM . For right-angle
triangles we need two pieces of information (other than the right-angle) to solve for a third.
Unfortunately we only have one piece of information for ∆KMN .
NSSAL
©2010
60
Draft
C. D. Pilmer
We do have three pieces of information for ∆JKN , so we should be able to solve for KN, the
side shared by ∆JKN and ∆KMN .
Once we have KN, we can use ∆KMN to solve for KM, and finally use ∆KLM to find ∠L
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
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°
NSSAL
©2010
61
Draft
C. D. Pilmer
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°
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°
NSSAL
©2010
62
Draft
C. D. Pilmer
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°
d 2 = 12100 + 6400 − 17600(0.9397 )
d 2 = 1961.28
d = 44.3 m
The ball is 44.3 m from the hole.
NSSAL
©2010
63
Draft
C. D. Pilmer
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.
NSSAL
©2010
64
Draft
C. D. Pilmer
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
NSSAL
©2010
65
Draft
C. D. Pilmer
(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
NSSAL
©2010
66
Draft
C. D. Pilmer
(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
NSSAL
©2010
67
Draft
C. D. Pilmer
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
NSSAL
©2010
9.6 cm
68
Draft
C. D. Pilmer
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
NSSAL
©2010
69
Draft
C. D. Pilmer
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?
NSSAL
©2010
70
Draft
C. D. Pilmer
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.
NSSAL
©2010
71
Draft
C. D. Pilmer
Putting It Together
1. Determine the indicated angle or side. Be prepared to use a variety of trigonometric tools
(i.e. similar triangles, 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
NSSAL
©2010
E
72
G
Draft
C. D. Pilmer
(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
NSSAL
©2010
73
Draft
C. D. Pilmer
2. Given the measures of DF and EF , find DE,
CD, and BD.
F
53.7 cm
44.1 cm
B
D
21.7 cm
E
C
3. Determine ∠EDA .
E
68o
D
C
47o
15.3 m
12.7 m
56o
A
NSSAL
©2010
18.2 m
B
74
Draft
C. D. Pilmer
4. Determine ML .
M
L
63o
12.7 m
57o
N
J
10.6 m
41o
13.9 m
K
5. 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?
6. How high is a tower that casts a 13.6 m shadow at the same time as a 1.6 metre high
individual casts a 1.1 metre shadow?
NSSAL
©2010
75
Draft
C. D. Pilmer
7. 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?
8. 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.
9. 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?
NSSAL
©2010
76
Draft
C. D. Pilmer
10. 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
11. 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
NSSAL
©2010
77
Second
Object
Draft
C. D. Pilmer
12. 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
13. 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.)
NSSAL
©2010
78
Draft
C. D. Pilmer
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?
NSSAL
©2010
79
Draft
C. D. Pilmer
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.
Prerequisite Practice (pages 1 to 4)
1. ∠DBC = 40° , ∠EBD = 140° , ∠ABC = 140° , ∠BDC = 45°
2. ∠ONP = 105° , ∠OPN = 35° , ∠NPQ = 145° , ∠QPR = 35°
3. ∠GHF = 55° , ∠HFC = 110° , ∠CFE = 70° , ∠DCF = 100° , ∠BCD = 80° , ∠EFG = 110°
4. ∠ILM = 115° , ∠MLN = 65° , ∠KLN = 115° , ∠KNL = 25° , ∠HIL = 115° , ∠JIL = 65°
Investigating Similar Triangles (pages 5 to 6)
Step 1: Triangle #1
∠CAB = 61°
∠ABC = 38°
∠BCA = 81°
Triangle #2
∠FDE = 61°
∠DEF = 38°
∠EFD = 81°
AB = 80 mm
BC = 70 mm
AC = 49 mm
DE = 122 mm
EF = 108 mm
DF = 75 mm
Corresponding angles are equal.
Step 2: Triangle #1
Triangle #2
Step 3:
AB 80
=
= 0.66
DE 122
BC 70
=
= 0.65
EF 108
AC 49
=
= 0.65
DF 75
All the ratios are equal.
Using Similar Triangles (pages 7 to 15)
1. PQ = 15.8 m
2. CD = 5.6 km
NSSAL
©2010
80
Draft
C. D. Pilmer
3. TU = 5.1 km
TV = 12.1 km
4. EF = 12.7 m
FG = 7.4 m
5. DE = 122 cm
6. LM = 54 m
JM = 128 m
7. RP = 47 cm
RQ = 26 cm
8. TV = 13.1 m
SV = 9.3 m
Similar Triangle Application Questions (pages 16 to 19)
1. 1.75 m
2. 57.6 m
3. 15.18 m
4. 0.41 m
5. 12 inches
The Pythagorean Theorem (pages 20 to 25)
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
NSSAL
©2010
81
Draft
C. D. Pilmer
Trigonometric Ratios (pages 26 to 29)
Investigation
Part 1
Triangle
Opposite
o
(20 )
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
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
Part 2
Triangle
(30o)
Part 3
Triangle
(40o)
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
NSSAL
©2010
82
Draft
C. D. Pilmer
Using Our Three Trigonometric Ratios, Part 1 (pages 30 to 35)
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 36 to 41)
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
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
NSSAL
©2010
83
Draft
C. D. Pilmer
Law of Sines (pages 42 to 45)
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 46 to 55)
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°
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
NSSAL
©2010
84
Draft
C. D. Pilmer
Law of Cosines (pages 56 to 58)
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.
Using the Law of Cosines (pages 59 to 71)
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
NSSAL
©2010
85
Draft
C. D. Pilmer
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 72 to 78)
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. DE = 30.6 cm, CD = 26.4 cm, BD = 15.1 cm
(Hint: Pythagorean Theorem and Similar Triangles)
3. First Step: BD = 17.4 m, Second Step: AD = 16.7 m, Final Step: ∠EDA = 58°
4. First Step: JM = 11.3 m, Second Step: JL = 10.2 m, Final Step: ML = 15.2 m
5. 134 m
6. 19.8 m
7. 2.6 km
8. (a) 10.9 m
(b) 9.3 m
9. 44.6 m
10. 8o
11. 221 m
12. (a) 130.5 m
(b) 27o
13. 811 m
NSSAL
©2010
86
Draft
C. D. Pilmer