Download Angles and Triangles Unit

Angles and Triangles Unit
(Level IV Graduate Math)
Draft
(NSSAL)
C. David Pilmer
©2014
(Last Update: April 2015)
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Table of Contents
Introduction………………………………………………………………………………… ii
Negotiated Completion Date………………………………………………………………. ii
The Big Picture…………………………………………………………………………….. iii
Course Timelines…………………………………………………………………………... iv
Angles and Degrees ………………………………………………………………………. 1
Relationships Amongst Angles …………………………………………………………… 9
Investigating Similar Triangles …………………………………………………………… 15
Using Similar Triangles …………………………………………………………………… 17
Similar Triangle Application Problems …………………………………………………… 26
The Pythagorean Theorem ………………………………………………………………… 30
Putting It Together ………………………………………………………………………… 37
Post-Unit Reflections………………………………………………………………………. 43
Soft Skills Rubric …………………………………………………………………………. 44
Answers……………………………………………………………………………………. 45
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Introduction
This unit has three components.
1. The first component focuses on measuring and estimating acute and obtuse angles in degree
measure. This is followed by understanding and exploiting relationships amongst angles
(e.g. supplementary angles, opposite angles, 180o in a triangle, etc.).
2. The second component of the unit focuses on using similar triangles to solve real world
problems.
3. The third and final component looks at real world applications and multi-step problems
involving the Pythagorean Theorem.
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 six 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. Instructors and students
may choose to alter this order to best serve the needs of the learner. Not all of the units take the same
amount of time to complete. For example, the Consumer Finance Unit is quite short. By contrast, the
Graphs and Function Unit and the Measurement Units (A and B) tend to take the greatest amount of time.
Math in the Real World Unit (Required)
 Fractions, decimals, percentages, ratios, proportions, and signed
numbers in real world applications
 Math Games and Puzzles
Solving Equations Unit (Required)
 Solve and check equations of the form
, and
.
,
Consumer Finance Unit (Required)
 Simple Interest and Compound Interest
 TVM Solver (Loans and Investments)
Graphs and Functions Unit (Required)
 Understanding Graphs
 Linear Functions and Line of Best Fit
Measurement Unit (Required)
 Part A: Imperial and Metric Measures
 Part B: Perimeter, Area and Volume
Angles and Triangles Unit (Required)
 Angle and Line Relationships
 Similar Triangles
 Pythagorean Theorem
Choose two of the four.
Linear
Functions and
Systems of
Equations Unit
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Trigonometry
Unit
iii
Descriptive
Statistics Unit
Numeracy Unit
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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
15
25
22
14
18
18
Total: 156 hours
Math in the Real World Unit
Solving Equations Unit
Consumer Finance Unit
Graphs and Functions Unit
Measurement Unit (Parts A & B)
Angles and Triangles Unit
Selected Unit #1
Selected Unit #2
Maximum
Completion Time
in Hours
34
28
18
30
30
16
22
22
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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Angles and Degrees
Aerial tricks done by skiers, figure skaters, and skateboarders are often
described using degrees. If the skier does a complete rotation off a jump, she is
said to have done a 360, which means a 360 degree rotation. Similarly, if the
skier does two complete rotations off a jump, she is said to have done a 720,
which means a 720 degree rotation.
In this section, we are going to learn how angles (or rotations) are measured in degrees.
An angle is:
 Formed by two line segments that start at the same point,
called a vertex.
 Measured in degrees, whose symbol is o.
vertex
30o
Angles that measure less than 90o are called acute angles.
70o
18o
40o
Angles that measure are between 90o and 180o are called obtuse angles.
107o
140o
153o
The degree measure of an angle is often determined using a protractor. The vertex of the angle is
placed on the crosshairs of the protractor, and one of the line segments is placed so that it is
along the horizontal base line of the protractor. Most protractors have an inner and outer scale;
use the one that is most appropriate for the angle you are measuring.
113o
55o
150o
28o
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Questions
1. Estimate the measure of the following angles. Also state whether this is an acute or obtuse
angle.
(a)
(b)
(c)
(d)
?
?
?
?
2. Determine the measure of the following angles. Also state whether this is an acute or obtuse
angle.
(a)
(b)
(c)
(d)
(e)
(f)
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3. Use a protractor to determine the degree measures of these angles.
(a)
(b)
?
?
(c)
(d)
?
?
(e)
(f)
?
?
(g)
(h)
?
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(i)
(j)
?
?
(k)
(l)
?
?
(m)
(n)
?
?
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Relationships Amongst Angles
Before looking at triangles, we need to know a few things about angles.
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
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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 .
A
C
B
E
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.
D
A
C
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).
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
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A
B
H
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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
1. Corresponding angles are equal.
FED  EIH , DEI  HIJ , FEG  EIK , and
GEI  KIJ
K
J
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
Questions
Find the identified angles. (Diagrams are not to scale.)
1.
2.
A
D
P
N
40o
30
o
C
B
70o
ACD 
O
NPO 
3.
4.
J
F
I
75
o
P
110
G
M
H
JIG 
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o
82o
O
N
PMO 
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5.
6.
G
F
V
U
o
37
W
o
D
105
R
E
T
S
DEG 
SVW 
7.
8.
A
D
Q
P
E
150o
50o
B
C
N
95o
M
AED 
ONP 
DEC 
OPN 
9.
O
10.
N
F
I
G
35o
58o
83o
H
P
M
O
FGH 
NMO 
FIH 
NOP 
11.
12.
E
o
o
32
S
D
49
100o
110
V
F
65o
U
T
G
FDG 
RUT 
DGF 
RUV 
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13.
14.
C
M
47o
A
o
N
B
81
D
Q
P
H
37o
E
F
R
O
G
CBD 
ONP 
FBD 
OPN 
BFH 
NPQ 
EFG 
OPR 
HFG 
QPR 
15.
16.
L
N
o
50
K
M
L
98o
60o
70o
H
P
65o
I
ONP 
NIH 
NMP 
NIJ 
LMN 
NMJ 
LKM 
MJK 
LMK 
LMJ 
KMP 
ONI 
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N
O
O
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17.
18.
B
C
G
H
H
40o
115o
D 95o
o
70
o
95
E
F
I
L
58o
N
J
G
GHF 
ILM 
HFC 
MLN 
CFE 
KLN 
DCF 
KNL 
BCD 
HIL 
EFG 
JIL 
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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?
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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?
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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
and
B
E
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D
C
are similar triangles.
Why?
1.
because they are the same
angle.
2.
because they are both labeled
as right angles.
3.
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.
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Diagram 2
and
are similar triangles.
S
R
Q
T
P
Why?
1.
because they are opposite
angles.
2.
because they are both labeled
as right angles.
3.
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
and
F
G
H
are similar triangles.
Why?
1.
because they are the same
angle.
2.
because when dealing with
parallel lines cut by a transversal line,
corresponding angles are equal.
3.
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
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31.7 cm
22.3 cm
C
E
43.5 cm
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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
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).
43.5c  706.91
706.91
43.5
c = 16.3 cm
c
Example 2
If JKL and MNO are similar triangles, find NO given the information supplied in the
diagram.
O
L
12.3 m
18.9 m
20.1 m
J
K
Answer:
JL
KL

MO NO
12.3 20.1

18.9
m
12.3m  20.1  18.9
12.3m  379.89
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M
N
Rather than calling our unknown side NO, we can call
it side m because it is opposite (or across from)
.
Often people will use the variable x to represent the
unknown side. Feel free to do that if you find it easier.
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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
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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

GH
71.8
51.2GH  168.3  71.8
51.2GH  12083.94
12083.94
51.2
GH = 236.0 cm
GH 
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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.
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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
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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
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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
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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
W
U
V
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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
h
2. 3
h  8.1 m
The height of the tree is 8.1 m.
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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.
h
1.5 m
3.5 m
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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
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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?
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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.
c = 7.66 cm
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
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3.9 2  2.32  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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(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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7. A wire is stretched from the top of a 4.2 metre antenna to a bracket 3.5 metres from the base
of the antenna. How long is the wire? Assume that the antenna is vertical and situated on a
horizontal flat piece of property.
8. Tylena has let out 50 metres of kite string when she observes that her kite is directly above
Candice. If Candice is 34 metres from Tylena, how high is the kite? Assume the kite string
is taunt and that the two women are standing on flat horizontal ground.
9. Two planes left the airport. One travelled north at 220 kilometres per hour. The other
travelled west at 260 kilometres per hour. How far apart will the planes be in a half hour?
10. The formula for the area of a parallelogram is A  bh , where b is the base and h is the
height. Determine the area of the parallelogram shown on the right.
35 cm
26 cm
15 cm
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Putting It Together
1. Determine the measure of the following angles. Also state whether this is an acute or obtuse
angle.
(a)
(b)
(c)
(d)
2. Use a protractor to determine the degree measures of these angles.
(a)
(b)
?
?
(c)
(d)
?
?
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(e)
(f)
?
?
(g)
(h)
?
?
3. In each case, find the indicated angles.
(a)
(b)
B
M
N
C
A
D
o
58
I
H
G
R
o
142
F
o
85
E
RPM 
ACH 
MPO 
GFC 
PMN 
GCF 
MNO 
DCF 
OPQ 
38
P
60
o
Q
O
IGC 
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(c)
(d)
P
C
A
B
65
Q
o
80
G
55o
N
D
H
F
M
80o O
o
L
K
I
43o
J
E
FEG 
NOI 
BGE 
QOI 
ABG 
MNK 
DBG 
LIJ 
DEF 
HKJ 
DEG 
IJK 
4. 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
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5. Calculate the length of GF.
G
7.6 m
D
F
E
3.3
m
5.7
m
6. If AB = 18.6 cm, EC = 10.4 cm,
AD = 12.4 cm, find ED, BD, CD,
and BC.
A
E
B
C
D
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7. Hinto initially walks diagonally across a rectangular property that measures 35 metres by 20
metres. After reaching the corner of the property, he decides to return to his starting position
by walking round the edge of the property. How much further does he walk on his return
trip?
8. 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?
9. Calculate the height of an equilateral triangle whose side length is 8 cm? (Note: For an
equilateral triangle, all sides are of equal length.)
h
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10. Jorell places a small mirror on the ground, 6.4 metres from the base of a tree. He then walks
backwards until he can see the top of the tree in the mirror. At this point, Jorell is 1.2 metres
away from the mirror. If his eye level is 1.5 metres, how tall is the tree? Assume this taking
place on flat horizontal ground.
11. A 6.8 metre rope is tied to the top of a 5.2 metre pole. The rope is pulled tight, extended its
full length, and stacked into the ground. How far is the staking point from the base of the
pole? Assume the pole is vertical and situated on flat horizonatl ground.
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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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Soft Skills Rubric
Look back over the module you have just completed and assess yourself using the following
rubric. Use pencil or pen and put a checkmark in the column that you think best describes your
competency for each description. I will look at how accurately you have done this and will
discuss with you any areas for improvement.
You will be better prepared for your next step, whether it is work or further education, if you are
competent in these areas by the end of the course. Keep all of these rubrics in one place and
check for improvement as you progress through the course.
Date:
Competent
demonstrates
concept
fully
consistently
Throughout this module, I…
the
and
Approaching
Competency
demonstrates the
concept most of the time
Developing
Competency
demonstrates the
concept some of the
time
 Attended every class
 Let my instructor know if not
able to attend class
 Arrived on time for class
 Took necessary materials to
class
 Used appropriate language for
class
 Used class time effectively
 Sustained commitment
throughout the module
 Persevered with tasks despite
difficulties
 Asked for help when needed
 Offered support/help to others
 Helped to maintain a positive
classroom environment
 Completed the module
according to negotiated
timeline
 Worked effectively without
close supervision
Comments:
(Created by Alice Veenema, Kingstec Campus)
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Answers
Angles and Degrees (pages 1 to 8)
1. (a)
(b)
(c)
(d)
acute; any value between or equal to 70o and 80o
obtuse; any value between or equal to 130o and 150o
acute; any value between or equal to 15o and 30o
obtuse; any value between or equal to 150o and 170o
2. (a) 100o, obtuse
(c) 145o, obtuse
(e) 14o, acute
(b) 60o, acute
(d) 42o, acute
(f) 168o, obtuse
3. If you are off by 2o or less, don't worry about it.
(a) 50o
(b)
(c) 150o
(d)
o
(e) 28
(f)
(g) 75o
(h)
(i) 95o
(j)
o
(k) 118
(l)
(m) 53o
(n)
34o
123o
113o
148o
12o
98o
85o
Relationships Amongst Angles (pages 9 to 14)
1.
ACD  140
2.
NPO  80
3.
JIG  75
4.
PMO  78
5.
DEG  53
6.
SVW  105
7.
AED  150
DEC  30
8.
ONP  50
OPN  35
9.
FGH  62
FIH  145
10.
NMO  32
NOP  90
11.
FDG  41
DGF  58
12.
RUT  85
RUV  95
13.
CBD  133
FBD  47
BFH  47
EFG  47
HFG  133
14.
ONP  99
OPN  44
NPQ  136
OPR  136
QPR  44
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15.
ONP  120
NMP  50
LMN  130
LKM  80
LMK  50
KMP  130
16.
NIH  98
NIJ  82
NMJ  125
MJK  125
LMJ  65
ONI  82
17.
GHF  52
HFC  110
CFE  70
DCF  100
BCD  80
EFG  110
18.
ILM  115
MLN  65
KLN  115
KNL  25
HIL  115
JIL  65
Investigating Similar Triangles (pages 15 and 16)
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 17 to 25)
1. PQ = 15.8 m
2. CD = 5.6 km
3. TU = 5.1 km
TV = 12.1 km
4. EF = 12.7 m
FG = 7.4 m
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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 26 to 29)
1. 1.75 m
2. 57.6 m
3. 15.18 m
4. 0.41 m
5. 12 inches
The Pythagorean Theorem (pages 30 to 36)
1. (a) 14.0 cm
(c) 14.8 cm
(b) 13.5 m
(d) 24.2 ft
2. It is right-angled because 82  152  17 2
3. (a) 10.6
(c) 10.8
(b) 9.8
4. 6.8 m
5. 11.3 m
6. 1.9 km
7. 5.5 m
8. 36.6 m
9. 170.3 km
10. 743 cm2
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Putting It Together (pages 37 to 42)
1. (a) 143o, obtuse
(c) 107o, obtuse
2. (a)
(c)
(e)
(g)
49o
155o
78o
65o
(b) 17o, acute
(d) 55o, acute
(b)
(d)
(f)
(h)
3.
(a) IGC  122
30o
115o
120o
156o
(b) RPM  120
ACH  58
MPO  60
GFC  38
PMN  135
GCF  84
MNO  80
DCF  38
OPQ  120
(c) FEG  47
(d) NOI  80
BGE  90
QOI  100
ABG  65
MNK  125
DBG  115
LIJ  80
DEF  122
HKJ  55
DEG  75
IJK  45
4. Hint: Involves both the Pythagorean Theorem and similar triangles.
DE = 30.6 cm, DC = 26.4 cm, BD = 15.0 cm
5. Hint: First find GE.
GF = 6.0 m
6. ED = 6.9 cm, DC = 12.5 cm, BD = 22.4 cm, BC = 9.9 cm
7. 55 – 40.3 = 14.7 m
8. 19.8 m
9. 6.9 cm
10. 8 m
11. 4.4 m
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