MFM2P
Foundations of Mathematics
Grade 10 Applied
Version A
MFM2P – Foundations of Mathematics
Introduction
Grade 10 Mathematics (Applied)
Welcome to the Grade 10 Foundations of Mathematics, MFM2P. This full-credit course
is part of the new Ontario Secondary School curriculum.
This course enables students to develop and understanding of the mathematical
concepts related to introductory algebra, proportional reasoning, and measurement and
geometry through investigation and the effective use of technology. Students will
investigate real-life examples to develop various representations of linear relationships,
and will determine the connections between the representations. They will also explore
certain relationships that emerge from the measurement of three-dimensional figures
and two-dimensional shapes. Students will consolidate their mathematical skills as they
solve problems and communicate their thinking.
Materials
This course is self-contained and does not require a textbook. You will require lined
paper, graph paper, a ruler, a scientific calculator and a writing utensil.
Expectations
The overall expectations you will cover in the lesson are listed on the first page of each
lesson.
Lesson Description
Each lesson contains one or more concepts with each being followed by support
questions. At the end of the lesson the key questions covering all concepts in the
lesson are assigned and will be submitted for evaluation.
Evaluation
In each lesson, there are support questions and key questions. You will be evaluated
on your answers to the key questions in each lesson, the mid-term exam and the final
exam.
Support Questions
These questions will help you understand the ideas and master the skills in each
lesson. They will also help you improve the way you communicate your ideas. The
support questions will prepare you for the key questions.
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MFM2P – Foundations of Mathematics
Introduction
Write your answers to the support questions in your notebook. Do not submit these
answers for evaluation. You can check your answers against the suggested answers
that are given at the end of each unit.
Key Questions
The key questions evaluate your achievement of the expectations for the lesson. Your
answers will show how well you have understood the ideas and mastered the skills.
They will also show how well you communicate your ideas.
You must try all the key questions and complete most of them successfully in order to
pass each unit. Write your answers to the key questions on your own paper and submit
them for evaluation at the end of each unit. Make sure each lesson number and
question is clearly labeled on your submitted work.
Mid-term and Final Examination
In this course there is a mid-term and final exam. These exams will incorporate the four
learning categories knowledge and understanding, application, communication, and
thinking and inquiry.
What You Must Do To Get a Credit
In order to be granted a credit in this course, you must
;
;
;
Successfully complete the Key Questions for each unit and submit them
for evaluation within the required time frame.
Complete the mid-term exam after Unit 2.
Complete and pass a final examination.
Your Final Mark
•
Each Unit has 5 lessons each worth 2% (10% per Unit x 4 Units)
Midterm Test
40%
30%
•
Final Examination
30%
•
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Term
Page 3 of 52
MFM2P – Foundations of Mathematics
Introduction
Unit One
Lesson One
¾ Similar triangle notation
¾ Interior angle property of similar triangles
¾ Proportionate side property of similar triangle
Lesson Two
¾ Triangle notation
¾ Substitution into the Pythagorean theorem
¾ Using Pythagorean theorem to find the missing side of a triangle
Lesson Three
¾
¾
¾
¾
Label parts of a right triangle i.e. opposite, adjacent and hypotenuse
Recognize parts of a right triangle i.e. opposite, adjacent and hypotenuse
Using the sine ratio to find unknown sides of a right triangle
Using the sine ratio to find unknown interior angles of a right triangle
Lesson Four
¾
¾
¾
¾
Label parts of a right triangle i.e. opposite, adjacent and hypotenuse
Recognize parts of a right triangle i.e. opposite, adjacent and hypotenuse
Using the cosine ratio to find unknown sides of a right triangle
Using the cosine ratio to find unknown interior angles of a right triangle
Lesson Five
¾
¾
¾
¾
Label parts of a right triangle i.e. opposite, adjacent and hypotenuse
Recognize parts of a right triangle i.e. opposite, adjacent and hypotenuse
Using the tangent ratio to find unknown sides of a right triangle
Using the tangent ratio to find unknown interior angles of a right triangle
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MFM2P – Foundations of Mathematics
Introduction
Unit Two
Lesson Six
¾ Explain and use correctly prefixes in the imperial and metric system
¾ Convert between imperial and metric units commonly used in everyday applications
Lesson Seven
¾ Radius and diameter
¾ Calculations using pi (π)
¾ Solving volume questions using formulas and substitution
Lesson Eight
¾
¾
¾
¾
Introduction to surface area
Radius and diameter
Calculations using pi (π)
Solving surface area questions using formulas and substitution
Lesson Nine
¾ Continued introduction to algebra
¾ Solving for unknowns
¾ Checking solutions to algebraic equations
Lesson Ten
¾ Using algebra to convert y-intercept form from standard form
¾ Using algebra to convert standard form from y-intercept form
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MFM2P – Foundations of Mathematics
Introduction
Unit Three
Lesson Eleven
¾
¾
¾
¾
¾
¾
¾
Introduction to the line
Using standard form of an equation
Using y-intercept form of an equation
x and y intercept
Recognizing positive, negative, zero and undefined slopes
Using the rise and the run of a given line to find its slope
Using a pair of coordinates of a line to calculate slope
Lesson Twelve
¾ Converting equation of a line from y-intercept form to standard form
¾ Recognizing and graphing parallel slopes
¾ Recognizing and graphing perpendicular slopes
Lesson Thirteen
¾
¾
¾
¾
Graphing linear equations
Checking whether a coordinate pair satisfies an equation
Finding the coordinates of the point of intersection of two equations
Recognizing the meaning of the point of intersection of a linear system of equations
Lesson Fourteen
¾
¾
¾
¾
Isolating a variable
Checking whether a coordinate pair satisfies an equation
Finding the coordinates of the point of intersection of two equations using algebra
Recognizing the meaning of the point of intersection of a linear system of equations
Lesson Fifteen
¾
¾
¾
¾
Solving system of linear equations using elimination
Checking whether a coordinate pair satisfies an equation
Finding the coordinates of the point of intersection of two equations using algebra
Recognizing the meaning of the point of intersection of a linear system of equations
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MFM2P – Foundations of Mathematics
Introduction
Unit Four
Lesson Sixteen
¾
¾
¾
¾
Collecting like terms
Distributive law
Expanding second degree polynomial expressions
Simplifying second degree polynomial expressions
Lesson Seventeen
¾ Finding the greatest common factor
¾ Dividing polynomials
Lesson Eighteen
¾ Factoring quadratic relations of the form ax 2 + bx + c where a = 1
¾ Factoring difference of squares trinomials
¾ Factoring perfect square trinomials
Lesson Nineteen
¾
¾
¾
¾
¾
Recognizing the direction of the opening of a parabola
Recognizing the values of the maximum and minimum
Recognizing the values of the x and y intercepts
Recognizing the coordinates of the vertex
Stating the equation of the axis of symmetry
Lesson Twenty
¾
¾
¾
¾
¾
Understanding the meaning of “a” in the equation y = a( x − h)2 + k
Understanding the meaning of “h” and “k” in the equation y = a( x − h)2 + k
Finding the axis of symmetry given the an equation in vertex form y = a( x − h)2 + k
Plotting and graphing quadratic equations
Substitution into quadratic equations
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Similar Triangles
Lesson 1
MFM2P – Foundations of Mathematics
Unit 1 – Lesson 1
Lesson One Concepts
¾ Similar triangle notation
¾ Interior angle property of similar triangles
¾ Proportionate side property of similar triangle
Similar Triangles
Similar triangles are triangles that have the same shape but not necessarily the same
size.
For a two triangles to be considered similar either one of two properties must apply.
This is the standard notation used for showing two triangles are similar to one another
ΔABC ≈ ΔXYZ. The order of the first three letters in not important, but what is important
is that each of the first three letters corresponds with its matching letter from the other
triangle.
Example 2
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MFM2P – Foundations of Mathematics
Unit 1 – Lesson 1
Prove that the following two triangles are similar and write the similarity statement .
Solution
In ΔEFG ∠G = 180 - 65 - 35 = 80° and in ΔSTU ∠U = 180 - 80 - 35 = 65°
Therefore both triangles have the same corresponding angles so
ΔEFG ≈ ΔTUS
Support Questions
1. For each of the following pairs of triangles explain whether or not they are similar.
Write a similarity statement for each pair of similar triangles.
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Page 10 of 52
MFM2P – Foundations of Mathematics
Unit 1 – Lesson 1
Support Questions
2. For each of the following pairs of triangles explain whether or not they are similar.
Write a similarity statement for each pair of similar triangles.
a.
b.
Property 2 The ratios of corresponding sides are equal.
Example 1
CA is the length
from C to A
AB BC CA
=
=
XY YZ ZX
AB 6
= =2
XY 3
BC 14
=
=2
YZ 7
CA 24
=
=2
ZX 12
or
XY 3
= = 0 .5
AB 6
YZ 7
=
= 0 .5
BC 14
ZX 12
=
= 0 .5
CA 24
Since the ratios of corresponding sides are the same then the triangles are similar.
ΔABC ≈ ΔXYZ
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MFM2P – Foundations of Mathematics
Unit 1 – Lesson 1
Example 2
Prove that the following two triangles are similar and write the similarity statement.
Solution
EF 25.5
=
=3
TS
8 .5
FG 12
=
=3
SU 4
GE 33
=
=3
UT 11
or
TS
8 .5
1
=
= = 0 .3 3
EF 25.5 3
SU 4 1
=
= = 0 .3 3
FG 12 3
UT 11 1
=
= = 0 .3 3
GE 33 3
Therefore both triangles have the same corresponding angles so
ΔEFG ≈ ΔTSU
Support Questions
3. For each of the following pairs of triangles explain whether or not they are similar. Write
a similarity statement for each pair of similar triangles.
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MFM2P – Foundations of Mathematics
Unit 1 – Lesson 1
Support Questions (continued)
Key Question #1
1. For each of the following pairs of triangles explain whether or not they are similar.
Write a similarity statement for each pair of similar triangles. (8 marks)
2. For each of the following pairs of triangles explain whether or not they are similar.
Write a similarity statement for each pair of similar triangles. (4 marks)
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MFM2P – Foundations of Mathematics
Unit 1 – Lesson 1
Key Question #1 (continued)
3. For each of the following pairs of triangles explain whether or not they are similar.
Write a similarity statement for each pair of similar triangles. (4 marks)
4. These triangles are the same shape yet have different sizes and orientations. Are
the triangles similar? Explain. (4 marks)
5. The following pairs of triangles are similar. Determine the value of x and y in each.
(6 marks)
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MFM2P – Foundations of Mathematics
Unit 1 – Lesson 1
Key Question #1 (continued)
6. In the diagram given below, a man 1.5 m tall is standing outside a church. How tall
is the church? (3 marks)
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The Pythagorean
Theorem
Lesson 2
MFM2P – Foundations of Mathematics
Unit 1 – Lesson 2
Lesson Two Concepts
¾ Triangle notation
¾ Substitution into the Pythagorean theorem
¾ Using Pythagorean theorem to find the missing side of a triangle
Pythagorean Theorem
The Pythagorean Theorem is one method used to calculate an unknown side of
a right angle triangle if the other two side are known.
Since the formula for the area of a square is A = s 2 then to find the length of the a side
of a square we square root the value of the area.
Example
Find the length of a side of a square that has an area of 100 cm 2 .
Solution
The area of square was 100 then the side would be 10.
We only use the
positive answer
since you cannot
have negative
length.
100 = ±10
Example
a. Find the length of the hypotenuse in the given triangle using Pythagorean Theorem.
3m
h
4m
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MFM2P – Foundations of Mathematics
Unit 1 – Lesson 2
b. Find the length of the missing side in the given triangle using Pythagorean Theorem.
x
18 cm
12 cm
Solution
a. Using Pythagorean Theorem a 2 + b 2 = c 2
a2 + b 2 = c 2
32 + 42 = c 2
25 = c 2
5=c
The area of square created by the
hypotenuse is 25 so the length of
its side is the square root of 25.
Therefore, the length of the hypotenuse is 5 m.
b. Using Pythagorean Theorem a 2 + b 2 = c 2
a2 + b 2 = c 2
x 2 + 122 = 182
x 2 = 182 − 122
x 2 = 180
x ≈ 13.4cm
Therefore, the length of the missing side is 13.4 cm.
Support Questions
1. Calculate the length of the third side of each triangle. Round to one decimal place.
a.
b.
7m
12 m
10.2 cm
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14.7 cm
Page 18 of 52
MFM2P – Foundations of Mathematics
Unit 1 – Lesson 2
Support Questions
2. Calculate the diagonal of each rectangle. Round to one decimal place.
a.
8 cm
14 cm
b.
1.5 m
2.7 m
3. The lengths of the sides of three triangles are given. Which are right triangles?
a. AB = 5 m, BC = 6.5 m, CA = 8.02 m
b. DE = 3 m, EF = 4 m, FD = 5 m
c. GH = 7.6 m, HI = 3.6 m, IG = 11.0 m
Key Question #2
1. What are the square roots of each number? (4 marks)
a. 14
c. 100
d. 1
1
36
e. .625
f. 1.73
g.
h.
25
64
2. Calculate the length of the third side of each triangle. Round to one decimal place.
(4 marks)
a.
b. 81
b.
5m
12 m
10.1 cm
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18.25 cm
Page 19 of 52
MFM2P – Foundations of Mathematics
Unit 1 – Lesson 2
Key Question #2 (continued)
3. The lengths of the sides of three triangles are given. Which are right triangles?
(6 marks)
a. AB = 24 m, BC = 10 m, CA = 26 m
b. DE = 7 m, EF = 8 m, FD = 13 m
c. GH = 10.6 m, HI = 5.6 m, IG = 9.0 m
4. A 7.5 m ladder is placed with its lower end 2 m away from the wall. How high up the
wall does the ladder reach? (3 marks)
5. Noah rows across a river that is 48 m wide. The current carries her 15 m
downstream. How far does Noah actually travel? (3 marks)
6. A new television measures 24” by 18”. What is the length of the diagonal of the
television? (3 marks)
7. The length of each rafter for a roof is 60 m including a 1 m overhang. The peak is
20.5 m above the horizontal beam. Find the width of the horizontal beam. (3 marks)
8. An electrical pole is 12.4 m tall. It is supported by a guy wire that is 17.3 m long.
How far is the anchor of the guy wire to the base of the electrical pole? (4 marks)
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Sine Ratio
Lesson 3
MFM2P – Foundations of Mathematics
Unit 1 – Lesson 3
Lesson Three Concepts
¾
¾
¾
¾
Label parts of a right triangle i.e. opposite, adjacent and hypotenuse
Recognize parts of a right triangle i.e. opposite, adjacent and hypotenuse
Using the sine ratio to find unknown sides of a right triangle
Using the sine ratio to find unknown interior angles of a right triangle
Trigonometry: The Sine Ratio
The sine ratio can be used calculate an unknown angle or an unknown side in a right
triangle.
To find the either of these unknowns the opposite side and hypotenuse are used from
an angle of reference.
Example
Suppose we use ∠A as the angle of reference then:
Sine always used the hypotenuse and opposite sides of a right triangle.
Sine Opposite Hypotenuse: SOH
When ∠ A is an acute angle in a right triangle, then
Sin A =
Length of side Opposite angle A
Length of Hypotenuse side
Sin A =
Opposite
Hypotenuse
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MFM2P – Foundations of Mathematics
Unit 1 – Lesson 3
How to find the missing side of a right triangle using the sine ratio.
Example 1
Find the value of x.
Solution
Sin A =
O
H
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MFM2P – Foundations of Mathematics
Unit 1 – Lesson 3
Support Questions
1. In each triangle, name the side:
a. opposite ∠E
b. the hypotenuse
2. Calculate the Sin A and Sin B in each triangle.
a.
b.
3. Calculate.
a. Sin 35°
b. Sin 71°
c. Sin 55°
d. Sin 90°
4. Calculate the value of x.
a.
b.
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MFM2P – Foundations of Mathematics
Unit 1 – Lesson 3
Support Questions (continued)
c.
d.
How to find the missing angle of a right triangle using the sine ratio.
Example 2
Find ∠A.
Solution
Sin A =
O
H
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MFM2P – Foundations of Mathematics
Unit 1 – Lesson 3
Support Questions
5. Calculate.
a. Sin −1 0.725
b. Sin −1 0.325
c. Sin −1
3
7
d. Sin −1
5
12
6. Calculate ∠E to the nearest degree.
a. Sin E = 0.625
b. Sin E = 0. 812
c. Sin E =
3
5
d. Sin E =
7
11
7. Calculate ∠E.
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Page 26 of 52
MFM2P – Foundations of Mathematics
Unit 1 – Lesson 3
Key Question #3
1. In each triangle, name the side: (2 marks)
2. Calculate the Sin A and Sin B in each triangle. (4 marks)
3. Calculate. (4 marks)
a. Sin 42°
b. Sin 68°
c. Sin 12°
d. Sin 0°
4. Calculate the value of x. (4 marks)
a.
b.
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MFM2P – Foundations of Mathematics
Unit 1 – Lesson 3
Key Question #3 (continued)
5. Calculate. (4 marks)
a. Sin −1 0.612
b. Sin −1 0.825
c. Sin −1
2
5
d. Sin −1
3
13
6. Calculate ∠E to the nearest degree. (4 marks)
a. Sin E = 0.387
b. Sin E = 0. 900
c. Sin E =
12
29
d. Sin E =
5
13
7. Calculate ∠ D. (8 marks)
8. A guy wire is 13.5 m long. It supports a vertical power pole. The wire is fastened to
the ground 9.5 m from the base of the 8.7 m tall pole. Calculate the measure of the
angle between the guy wire and the ground. (3 marks)
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MFM2P – Foundations of Mathematics
Unit 1 – Lesson 3
Key Question #3 (continued)
9. A 5.0 m ladder is leaning 3.7 m up a wall. What is the angle the ladder
makes with the ground? (3 marks)
10. A 12 m ladder leaning up against a wall makes a 50° angle with the
ground. How far up the wall does the ladder reach? (3 marks)
5
is an error. Draw and label a triangle with the known side
3
lengths given in the ratio and then explain why the error occurs. (4 marks)
11. Explain why Sin A =
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Page 29 of 52
Cosine Ratio
Lesson 4
MFM2P – Foundations of Mathematics
Unit 1 – Lesson 4
Lesson Four Concepts
¾
¾
¾
¾
Label parts of a right triangle i.e. opposite, adjacent and hypotenuse
Recognize parts of a right triangle i.e. opposite, adjacent and hypotenuse
Using the cosine ratio to find unknown sides of a right triangle
Using the cosine ratio to find unknown interior angles of a right triangle
Trigonometry: The Cosine Ratio
The cosine ratio can be used calculate an unknown angle or and unknown side in a
right triangle.
To find the either of these unknowns the adjacent side and hypotenuse are used from
an angle of reference.
Example
Suppose we use ∠A as the angle of reference then:
Cosine always used the hypotenuse and adjacent sides of a right triangle.
Cosine Adjacent Hypotenuse: CAH
When ∠ A is an acute angle in a right triangle, then
Cos A =
Length of side Adjacent angle A
Length of Hypotenuse side
Cos A =
Adjacent
Hypotenuse
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Page 31 of 52
MFM2P – Foundations of Mathematics
Unit 1 – Lesson 4
How to find the missing side of a right triangle using the Cosine ratio.
Example 1
Find the value of x.
Solution
Cos A =
A
H
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MFM2P – Foundations of Mathematics
Unit 1 – Lesson 4
Support Questions
1. In each triangle, name the side:
a. adjacent ∠ E
b. adjacent ∠ Y
2. Calculate the Cos A and Cos B in each triangle.
a.
b.
3. Calculate.
a. Cos 35°
b. Cos 71°
c. Cos 55°
d. Cos 90°
4. Calculate the value of x.
a.
b.
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MFM2P – Foundations of Mathematics
Unit 1 – Lesson 4
Support Questions (continued)
c.
d.
How to find the missing angle of a right triangle using the Cosine ratio.
Example 2
Find ∠C.
Solution
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MFM2P – Foundations of Mathematics
Unit 1 – Lesson 4
Support Questions
5. Calculate.
a. Cos −1 0.725
b. Cos −1 0.325
c. Cos −1
3
7
d. Cos −1
5
12
6. Calculate ∠E to the nearest degree.
a. Cos E = 0.625
b. Cos E = 0. 812
c. Cos E =
3
5
d. Cos E =
7
11
7. Calculate ∠E.
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MFM2P – Foundations of Mathematics
Unit 1 – Lesson 4
Key Question #4
1. In each triangle, name the side: (2 marks)
2. Calculate the Cos A and Cos B in each triangle. (4 marks)
3. Calculate. (4 marks)
a. Cos 42°
b. Cos 68°
c. Cos 12°
d. Cos 0°
4. Calculate the value of x. (4 marks)
a.
b.
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MFM2P – Foundations of Mathematics
Unit 1 – Lesson 4
Key Question #4 (continued)
5. Calculate. (4 marks)
a. Cos −1 0.612
b. Cos −1 0.825
c. Cos −1
2
5
d. Cos −1
3
13
6. Calculate ∠E to the nearest degree. (4 marks)
a. Cos E = 0.387
b. Cos E = 0. 900 c. Cos E =
12
29
d. Cos E =
5
13
7. Calculate ∠E. (8 marks)
8. For safety, the angle between a ladder and the ground should be between 60 and
78. A ladder 8 m in length is placed 3 m from the base of a wall. Is it safe to climb
the ladder? (3 marks)
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MFM2P – Foundations of Mathematics
Unit 1 – Lesson 4
Key Question #4 (continued)
9. A kite has a string 100 m long anchored to the ground. The string makes and angle
with the ground of 68°. What is the horizontal distance of the kite from the anchor?
(3 marks)
10. A ladder is leaned against a wall with its base 6 m from the wall. The ladder makes
a 50° angle with the ground. How long is the ladder? (3 marks)
11. Does the cosine ratio work with non-right triangles? Explain and prove your answer
with an example. (4 marks)
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Page 38 of 52
Tangent Ratio
Lesson 5
MFM2P – Foundations of Mathematics
Unit 1 – Lesson 5
Lesson Five Concepts
¾
¾
¾
¾
Label parts of a right triangle i.e. opposite, adjacent and hypotenuse
Recognize parts of a right triangle i.e. opposite, adjacent and hypotenuse
Using the tangent ratio to find unknown sides of a right triangle
Using the tangent ratio to find unknown interior angles of a right triangle
Trigonometry: The Tangent Ratio
The Tangent ratio can also be used to calculate an unknown angle or unknown side in a
right triangle.
To find either of these unknowns the adjacent side and opposite sides are used from an
angle of reference.
Example
Suppose we use ∠A as the angle of reference then:
Tangent always used the opposite and adjacent sides of a right triangle.
Tangent Opposite Adjacent: TOA
When ∠ A is an acute angle in a right triangle, then
Tan A =
Length of side opposite angle A
Length of Adjacent side
Tan A =
Opposite
Adjacent
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Page 40 of 52
MFM2P – Foundations of Mathematics
Unit 1 – Lesson 5
How to find the missing side of a right triangle using the Tangent ratio.
Example 1
Find the value of x.
Solution
Tan A =
A
H
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MFM2P – Foundations of Mathematics
Unit 1 – Lesson 5
Support Questions
1. In each triangle, name the side:
a. adjacent ∠ F
b. opposite ∠ Y
2. Calculate the Tan A and Tan B in each triangle.
a.
b.
3. Calculate.
a. Tan 35°
b. Tan 71°
c. Tan 55°
d. Tan 90°
4. Calculate the value of x.
a.
b.
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MFM2P – Foundations of Mathematics
Unit 1 – Lesson 5
Support Questions (continued)
c.
d.
How to find the missing angle of a right triangle using the Cosine ratio.
Example 2
Find ∠A.
Solution
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Page 43 of 52
MFM2P – Foundations of Mathematics
Unit 1 – Lesson 5
Support Questions
5. Calculate.
a. Tan −1 0.725
b. Tan −1 0.32
c. Tan −1
3
7
d. Tan −1
5
12
6. Calculate ∠E to the nearest degree.
a. Tan E = 0.625
b. Tan E = 0. 812
c. Tan E =
3
5
d. Tan E =
7
11
7. Calculate ∠E.
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Page 44 of 52
MFM2P – Foundations of Mathematics
Unit 1 – Lesson 5
Key Question #5
1. In each triangle, name the side: (2 marks)
2. Calculate the Tan A and Tan B in each triangle. (4 marks)
3. Calculate. (4 marks)
a. Tan 42°
b. Tan 68°
c. Tan 12°
d. Tan 50°
4. Calculate the value of x. (4 marks)
a.
b.
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Page 45 of 52
MFM2P – Foundations of Mathematics
Unit 1 – Lesson 5
Key Question #5 (continued)
5. Calculate. (4 marks)
a. Tan −1 1.612
b. Tan −1 0.825
c. Tan −1
5
2
d. Tan −1
3
13
6. Calculate ∠E to the nearest degree. (4 marks)
a. Tan E = 0.387
b. Tan E = 2 .90
c. Tan E =
12
29
d. Tan E =
13
5
7. Calculate ∠E. (8 marks)
8. Building A and building B are 15 m apart. Building B is 75 m high. What is the angle
(angle of elevation) from the base of building A to the top of building B? (3 marks)
Copyright © 2005, Durham Continuing Education
Page 46 of 52
MFM2P – Foundations of Mathematics
Unit 1 – Lesson 5
Key Question #5 (continued)
9. The top of a communications tower is 150 m above sea level. From a boat at sea,
its angle of elevation is 3°. (5 marks)
a. Using the diagram given above, what is meant by the angle of elevation?
b. How far is the boat from the tower?
10. A ladder is leaned 10 m up a wall with its base 6 m from the wall. What angle does
the ladder make with the ground? (3 marks)
11. The acronym SOHCAHTOA is often used in trigonometry. What do you think each
letter stands for and give an example finding either an unknown side or an unknown
angle using a portion of this acronym. (4 marks)
Copyright © 2005, Durham Continuing Education
Page 47 of 52
MFM2P – Foundations of Mathematics
Support Question Answers
Answers to Support Questions
Lesson One
similar; all angles correspond; ΔDEF≈ΔABC
similar; all angles correspond; ΔQRS≈ΔTUV
not similar because don’t know if all angles correspond with each other
similar; all angles correspond; ΔABC≈ΔDEF
1.
a.
b.
c.
d.
2.
a. similar; all angles correspond; ΔABC≈ΔDEF
b. similar; all angles correspond; ΔQRS≈ΔXWY
2 4 6
a. similar; sides are proportionate; = = = 2
1 2 3
15 20 25
=
=
=5
b. similar; sides are proportionate;
3
4
5
2 3 6
c. not similar; sides are not proportionate; = ≠
4 6 9
9 15 35
≠
d. not similar; sides are not proportionate; =
3 5 12
3.
Lesson Two
1.
a.
b.
a 2 + (10.2)2 = (14.7)2
72 + 122 = c 2
a 2 + (10.2)2 − (10.2)2 = (14.7)2 − (10.2)2
49 + 144 = c 2
a 2 = (14.7)2 − (10.2)2
193 = c 2
13.9 ≈ c
a 2 = 216.09 − 104.04
a 2 = 112.05
a ≈ 10.6
2.
a.
b.
8 + 14 = c
2
(1.5)2 + (2.7)2 = c 2
64 + 196 = c 2
2.25 + 7.29 = c 2
260 = c 2
16.1 ≈ c
9.54 = c 2
3.1 ≈ c
2
2
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Page 48 of 52
MFM2P – Foundations of Mathematics
3.
Support Question Answers
a.
b.
52 + 6.52 = 8.022
25 + 42.25 = 64.3204
32 + 42 = 52
9 + 16 = 25
25 = 25
yes
67.25 ≠ 64.3204
no
c.
7.62 + 3.62 = 112
57.76 + 12.96 = 121
70.72 ≠ 121
no
Lesson Three
b. YZ
1.
a. FG
2.
a. SinA =
3.
a. 0.574
4.
a.
3.7
1.5
; SinB =
4
4
b. SinA =
b. 0.946
12
13.4
; SinB =
18
18
c. 0.819
d. 1
b.
x
Sin 37 =
3.7
x = 3.7 Sin 37
x = 2.23 cm
c.
x
18
x = 18 Sin 23
x = 7.03 cm
Sin 23 =
d.
4
x
4 = x Sin 48
x Sin 48
4
=
Sin 48
Sin 48
4
=x
Sin 48
5.28 = x
13.3
x
13.4 = x Sin 62
x Sin 62
13.4
=
Sin 62
Sin 62
13.4
=x
Sin 62
15.2 = x
Sin 48 =
5.
a. 46°
b. 19°
6.
Sin −1 0.625
a.
= 39°
Sin −1 0.812
b.
= 54°
Sin 62 =
c. 25°
c.
Copyright © 2005, Durham Continuing Education
Sin −1
= 37°
d. 25°
3
5
7
d.
11
= 40°
Sin −1
Page 49 of 52
MFM2P – Foundations of Mathematics
7.
a.
Support Question Answers
b.
SinE =
2.1
3.7
∠E = Sin −1
c.
SinE =
2.1
3.7
10
18
∠E = Sin −1
∠E = 35°
d.
SinE =
10
18
4
9
∠E = Sin −1
SinE =
4
9
∠E = Sin −1
∠E = 26°
∠E = 34°
13.4
16.1
13.4
16.1
∠E = 56°
Lesson Four
b. XY
1.
a. GE
2.
a. CosA =
3.
a. 0.819
4
a.
1.5
3.7
; CosB =
4
4
b. CosA =
b. 0.326
13.4
12
; CosB =
18
18
c. 0.574
d. 0
b.
x
3.7
x = 3.7 Cos 37
x = 2.95 cm
x
18
x = 18 Cos 23
x = 16.57 cm
Cos 23 =
Cos 37 =
c.
d.
4
x
4 = x Cos 48
4
x Cos 48
=
Cos 48
Cos 48
4
=x
Cos 48
5.98 = x
13.4
x
13.4 = x Cos 62
13.4
xCos 62
=
Cos 62
Cos 62
13.4
=x
Cos 62
28.5 = x
Cos 48 =
5.
a. 44°
6.
a.
Cos −10.725
= 43.5°
b. 71°
b.
Cos −10.325
= 71°
Cos 62 =
c. 65°
c.
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3
7
= 64.6°
Cos −1
d. 65°
d.
5
12
= 65.4°
Cos −1
Page 50 of 52
MFM2P – Foundations of Mathematics
7.
8.
a.
Cos −1 0.625
= 51°
a.
2.1
CosE =
3.7
∠E = Cos −1
2.1
3.7
∠E = 55°
b.
Support Question Answers
Cos −1 0.812
= 36°
c.
Cos −1
3
5
c.
10
CosE =
18
4
CosE =
9
10
18
7
11
d.
CosE =
∠E = Cos −1
∠E = 56°
Cos −1
= 50°
= 53°
b.
∠E = Cos −1
d.
4
9
13.4
16.1
∠E = Cos −1
∠E = 64°
13.4
16.1
∠E = 34°
Lesson Five
b. XZ
1.
a. FG
2.
a. TanA =
3.
a. 0.700
4.
a.
3.7
1.5
; TanB =
1.5
3.7
b. TanA =
b. 2.90
12
13.4
; TanB =
13.4
12
c. 1.43
d. undefined (error)
b.
18
x
18 = x Tan 23
18
x Tan 23
=
Tan 23
Tan 23
18
=x
Tan 23
42.4 = x
Tan 23 =
x
3.7
x = 3.7 Tan 37
x = 2.79 cm
Tan 37 =
c.
d.
x
Tan 48 =
4
x = 4 Tan 48
x = 4.44 m
5.
a. 36°
b. 18°
6.
Tan −1 0.625
a.
= 32°
Tan −1 0.812
b.
= 39°
x
13.4
x = 13.4 Tan 62
x = 25.2 cm
Tan 62 =
c. 23°
c.
Copyright © 2005, Durham Continuing Education
Tan −1
= 31°
d. 23°
3
5
d.
Tan −1
7
11
= 32°
Page 51 of 52
MFM2P – Foundations of Mathematics
7.
a.
Support Question Answers
b.
TanE =
3.7
2.1
∠E = Tan −1
∠E = 60°
c.
TanE =
3.7
2.1
18
10
∠E = Tan −1
d.
TanE =
18
10
∠E = 61°
Copyright © 2005, Durham Continuing Education
9
4
∠E = Tan −1
∠E = 66°
TanE =
9
4
16.1
13.4
∠E = Tan −1
16.1
13.4
∠E = 50°
Page 52 of 52