MATH 10 COMMON
Part 3
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Teacher:
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Table of Contents
UNIT 5: SYSTEMS OF EQUATIONS ................................................................................................................................2
UNIT 5 OUTLINE: SYSTEMS OF EQUATIONS .............................................................................................................................. 3
UNIT 5 STUDY GUIDE: SYSTEMS OF EQUATIONS ........................................................................................................................ 4
Lesson 1 – Developing Systems ................................................................................................................................. 7
Lesson 2 – Solving Graphically .................................................................................................................................. 10
Lesson 3 – Properties of Linear Systems .................................................................................................................... 13
Lesson 4 – Solving by Substitution............................................................................................................................. 16
Lesson 5 – Solving Word Problems by Substitution ..................................................................................................... 19
Lesson 6 – Solving by Elimination .............................................................................................................................. 22
Lesson 7 – Solving Word Problems by Elimination ...................................................................................................... 24
UNIT 6: TRIGONOMETRY ........................................................................................................................................... 26
UNIT 6 OUTLINE: TRIGONOMETRY ........................................................................................................................................ 27
UNIT 6 STUDY GUIDE: TRIGONOMETRY .................................................................................................................................. 28
Lesson 1 – Pythagorean Theorem .............................................................................................................................. 30
Lesson 2 – Solving for Angles .................................................................................................................................... 34
Lesson 3 – Solving for Side Lengths ........................................................................................................................... 38
Lesson 4 – Solving Triangles...................................................................................................................................... 41
Lesson 5 – Solving with Multiple Triangles .................................................................................................................. 43
Lesson 6 – Word Problems with Multiple Triangles...................................................................................................... 47
UNIT 7: MEASUREMENT ............................................................................................................................................ 49
UNIT 7 OUTLINE: MEASURMENT .......................................................................................................................................... 50
UNIT 7 STUDY GUIDE: MEASUREMENT ................................................................................................................................... 51
Lesson 1 – Imperial Measures of Length .................................................................................................................... 55
Lesson 2 – Converting between SI and Imperial ......................................................................................................... 58
Lesson 3 – Review: Pyramids and Cones ................................................................................................................... 62
Lesson 4 – Surface Area of 3D Shapes ....................................................................................................................... 66
Lesson 5 – Volume of 3D Shapes............................................................................................................................... 69
Lesson 6 – Composite Shapes .................................................................................................................................. 72
Unit 5: Systems of
Equations
Quiz:
S.G Quiz:
Unit Test:
Page | 2
Unit 5 Outline: Systems of Equations
Homework or practice problems are a necessity to understand and do well in mathematics. If you have di iculty
with your homework, read your notes, read the examples in the textbook, ask questions in class, and attend extra
help session before or after school.
Absent students are responsible for all assigned work, regardless of whether the absence is excused or not.
You do not have to do both A and B homework where indicated, complete A for the average assignment and B if you
are looking for a more of a challenge.
Day
Topic
Page/Source
Pg. 401
1
Developing Systems
2
Solving Graphically
Exercise 8-1
Worksheet
3
Properties of Linear Systems
Exercise 8-5
Worksheet
4
Quiz
5
Solve by Substitution
6
7
8
Substitution
Word Problems
Solve by Elimination
Solve by Elimination
Word Problems
Worksheet
Exercise 8.6
Assignment
# 4 – 9, 12
#1 – 7, Chose 10 from 8 – 20
A: #1, 2 – 4aceg
B: # 2 – 4aceg, 5
A: # 1 – 4
B: # 1 – 5
Date:
Exercise 8-4
Worksheet
A: # 1abc, 2, 3, 4aceg, 10 – 20
Exercise 8-7
Worksheet
A: #1 – 21 odd
Exercise 8-3
Worksheet
A: # 1 – 4, 5aceg, 6abc
Exercise 8-8
Exercise 8-9
Chose 10 from 8-8
9
Study Guide Quiz
Due:
10
Unit Exam
Date:
B: # 1 – 5, 10 – 20
B: # 9 – 22
B: # 1, 2&3aceg, 4 – 7
Chose 10 from 8-9
Page | 3
Unit 5 Study Guide: Systems of Equations
1. Without graphing, determine whether each system of equations has one solution, no solutions,
or infinitely many solutions. Include your reasons with respect to slope and y-intercept.
a)
b)
i) 𝑦 = 𝑥 + 3
ii) 𝑦 = 3𝑥 + 3
Number of Solutions:
Reasons:
i) 2𝑦 = 3𝑥 + 8
ii) 8𝑦 + 4 = 12𝑥
Number of Solutions:
Reasons:
2. A linear system is defined by: 2𝑥 + 2𝑦 = 10 and
a) Solve the system by graphing
c) i) 4𝑥 + 2𝑦 = 4
ii) 2𝑥 + 𝑦 = 2
Number of Solutions:
Reasons:
3 − 𝑦 − 3𝑥 = 0
b) Algebraically verify your solution.
Solution:
3. A linear system is defined by: 3𝑥 + 𝑦 = 13 and
a) Solve the system using the Substitution
Method
3𝑦 + 2𝑥 = 18
b) Algebraically verify your solution.
Solution:
4. A linear system is defined by:
𝑥 + 𝑦 = 3 and
a) Solve the system using the Elimination
Method
3𝑥 − 𝑦 = 5
b) Algebraically verify your solution.
Solution:
Page | 4
5. A linear system is defined by: 7𝑥 + 6 = 3𝑦 and
a) Solve the system using either Substitution
Method or Elimination Method
19 + 3𝑥 − 2𝑦 = 0
b) Algebraically verify your solution.
Solution:
6. The amount of vitamin C in one apple and two peaches it 17 mg. The amount of vitamin C in two
apples and one peach is 13 mg.
a) Define your variables and set up a system of two equations to model this scenario.
b) What is the mass (in mg) of Vitamin C in one apple and in one peach?
7. The difference between two number is 23. The larger number is one more than triple the
smaller.
a) Define your variables and set up two equations.
b) Verify that the solution to your system is 11 and 34.
8. A collection of quarters (25¢) and nickels (5¢) are worth $5.90. If there are 90 coins, how many
were nickels and how many were quarters? Fill in this chart and create two equations.
Number of Coins
Value of One Coin
Total Value
Quarters
Nickels
Total
Equations:
Solve your equations and find out how many nickels and how many quarters there were. Write a
sentence to explain your answer.
Page | 5
9. A bulk store combines raisins and peanuts to form a mixture. There is 24 lbs. of mixture which
they value at $2.05 per lb. The store creates the mixture from peanuts which are valued at $1.60
per lb. and raisins which are valued at $2.80 per pound. How many lbs. of raisins and how many
lbs. of peanuts were used? Fill in this chart and create two equations.
[1 mark]
Number of lbs.
Value of One lb.
Total Value
Peanuts
Raisins
Mixture
Equations:
[1 mark]
Page | 6
Lesson 1 – Developing Systems
REVIEW:
Which of the following are linear equa ons?
2
𝑦 = 𝑥+6
3
2𝑥 − 3𝑦 = 19
𝑦 =𝑥 +2
DEFINITIONS:
A system of linear equa ons is two linear func ons which use the same two variables (usually
x and y)
A solu on is a pair of values that sa sfy BOTH equa ons.
Example 1: Given the following systems of linear equa ons verify that the solu on (2, −5) is
correct.
2𝑥 + 𝑦 = −1
3𝑥 − 2𝑦 = 16
Example 2: Determine which of the following solu on pairs is a solu on for the given linear
system of equa ons.
𝑐 = 2𝑑 + 2
𝑐 + 2𝑑 = −6
Solu on A: 𝑐 = 4, 𝑑 = 1
Sou on B : 𝑐 = −2, 𝑑 = −2
Page | 7
Se ng up a Linear System of Equa ons:
1) Define the two different variables you are using and what they represent.
2) Based on the context, write two equa ons using those variables
Example 3: A school raised $195 by collec ng 3000 items for recycling. The school received 5
cents for each pop can and 20 cents for each large plas c bo le.
a) Define the variables.
b) Create a linear system for this situa on.
c) The school collected 2700 pop cans and 300 plas c bo les. Use the linear system to
verify these numbers.
Example 4: Fred bought 3 adult ckets and two student ckets to the school play for a total of
$31.00. Allison bought one adult cket and one student cket for a total of $12.00. Set up a
linear equa on system that represents this situa on.
a) Define the variables.
b) Create a linear system for this situa on.
c) Given that each adult cket is $7.00, how much is each student cket? Verify your
equa ons.
Page | 8
Example 5: Sam’s bike shop sells bicycles and tricycles. The total number of wheels on either
bicycles or tricycles in his store is 45. There are 10 more bicycles than tricycles in his store.
a) Define the variables.
b) Create a linear system for this situa on.
c) One possible solu on could be that there are 15 bicycles and 5 tricycles. Verify this
solu on.
Page | 9
Lesson 2 – Solving Graphically
REVIEW:
Graph each line by hand:
𝑦 = 2𝑥 − 6
3𝑥 + 4𝑦 = 9
At what point do they intersect?
What does that mean?
Example 1: Solve this linear system by graphing and sketch the graphs. Don’t forget to verify!
𝑥+𝑦=8
3𝑥 − 2𝑦 = 14
Page | 10
Example 2: Solve this linear system by graphing and sketch the graphs.
𝑥−𝑦 =4
2𝑥 − 3𝑦 = 3
SOLVING BY GRAPHING USING A CALCULATOR
When we solve systems of equa ons graphically we are looking for the point of intersec on.
Step 1) Isolate y in both equa ons
Step 2) Input equa ons into y1 and y2 on your calculator
Step 3) Find the point of intersec on: [2ND] [TRACE] 5: Intersec on
Step 4) Verify!
Example 3: Solve using graphically using technology.
5𝑦 = 2𝑥 + 15
3𝑥 + 𝑦 + 4 = 0
Page | 11
Example 4: Create a system of equa ons and solve graphically
To visit Head-Smashed-in Buffalo Jump Interpre ve center, the admission fee for a student is
$5.00 and $9.00 for an adult. In one hour, 32 people entered the center and a total of $180.00
was collected. How many students and how many adults visited during this me?
a) Define your variables.
b) Create a system of equa ons.
c) Solve graphically.
d) Verify your solu on.
Example 5: Create a system of equa ons and solve graphically.
Six pencils and four crayons cost $3.40. Three pencils and ten crayons cost $4.90. How much
does one pencil cost?
Page | 12
Lesson 3 – Properties of Linear Systems
REVIEW: Solve and sketch each system
2𝑥 + 𝑦 = 1
4𝑥 + 6𝑦 = −18
3𝑥 + 𝑦 = −2
−6𝑥 − 2𝑦 = 12
𝑥 + 𝑦 = −2
−2𝑥 − 2𝑦 = 4
Solu on:
Solu on:
Solu on:
What do we see?
Page | 13
Type of Lines
Intersec ng
Graph
Slopes
Intercepts
# of Solu ons
Coincident
Parallel
Example 1: Classify without graphing
a)
3𝑥 + 5𝑦 = 9
6𝑥 + 10𝑦 = 18
b)
𝑥 + 2𝑦 = 6
𝑥 + 2𝑦 = 2
c)
𝑥+𝑦 =5
3𝑥 + 𝑦 = 3
Page | 14
Example 2: Consider the equa on 3𝑥 − 6𝑦 = 12. Find another linear equa on that will make
the system have:
a) 1 Solu on
b) Infinite Solu ons
c) No Solu on
Example 3: Classify these equa ons by how many solu ons there will be, without graphing.
a)
3𝑥 + 𝑦 = 10
𝑥 − 2𝑦 = 1
b)
𝑥 − 2𝑦 = −6
𝑦 = 𝑥+3
c)
𝑦=− 𝑥
3𝑥 + 4𝑦 = 12
Page | 15
Lesson 4 – Solving by Substitution
Solving algebraically (by subs tu on or elimina on) is important
because we can find the EXACT solu on. Graphing, even with
calculators, can some mes only give us an approximate solu on.
Steps to Solve by Subs tu on:
1) Isolate a variable in one equa on
2) Replace the variable in the second equa on with the isolated equa on
3) Solve for the variable
4) Use the found value from step 3 to find the second variable using either original
equa on
5) Verify your solu on
6) Conclude with a solu on (point of intersec on)
Example 1: Prac ce Subs tu on 𝑦 = 𝑥 − 3𝑥 − 1
a) 𝑥 = −3
b) 𝑥 = 2𝑎
c) 𝑥 = 𝑦 + 1
Page | 16
Example 2: Choose a variable to isolate
3𝑥 − 𝑦 − 9 = 0
𝑥 + 2𝑦 = 7
1
𝑦 − 5 = 3𝑥
2
Example 3: Solve by Subs tu on
3𝑥 + 4𝑦 = −4
𝑥 + 2𝑦 = 2
Step 1) Isolate one variable in one equa on
Step 2) Subs tute
Step 3) Solve
Step 4) Subs tute and solve for the other variable
Step 5) Verify and State
Page | 17
Example 4: Solve by Subs tu on
2𝑥 − 4𝑦 = 7
4𝑥 + 𝑦 = 5
Example 5: Solve by Subs tu on
1
2
𝑥 + 𝑦 = −1
2
3
1
5
𝑦= 𝑥−
4
3
Page | 18
Lesson 5 – Solving Word Problems by Substitution
What is the value of each weight?
Approaching Word Problems:
1. Define variables with a let statement
2. Develop a System of equa ons
3. Solve
4. Verify
5. Closing Sentence
Example 1: During a sale, all shirts are on sale for one price and all sweaters are on sale for
another price. Mario bought two shirts and four sweaters for $190.00. Jordan bought one shirt
and three sweaters for $135.00. Determine the sale price for each item.
Page | 19
Example 2: Billy has 14 coins which are a collec on of dimes and quarters in his pocket. If the
total value of $3.05, how many of each kind of coin does he have?
Coin Definitions
Nickel (5 cents) $0.05
Dime (10 cents) $0.10
Quarter (25 cents) $0.25
Loonie (100 cents) $1.00
Toonie (200 cents) $2.00
Example 3: The difference between two numbers is 7. Two mes the larger of the two
numbers is equal to the sum of five and three mes the smaller number. Find the two
numbers.
Page | 20
Example 4: A movie theatre is coun ng their profit for the day. They charge $12.00 for adults
and $8.00 for children. Over the course of the day, they made a profit of $9056.00. If 858
ckets were bought, how many adults and children went to the movies that day?
Example 5: Sandra invests $1670. Some of it goes into a GIC at 2.5% interest and the other
por on goes in a RSP at 3%. Both are compounded annually. A er one year the interest earned
is $46.85. How much money did Sandra ini ally invest at the start of the year in each account?
Page | 21
Lesson 6 – Solving by Elimination
How to Solve Using Elimina on
1. “Stack” each like term.
2. Does at least one variable have the same coefficients?
Yes -> Cancel out the variable that matches using addi on or subtrac on
No -> Mul ply one equa on so that the coefficients for one variable are the same
3. Subs tute the value you found into either equa on to find the missing variable
4. Verify! And state your solu on
Example 1: Solve by elimina on
10 = 𝑥 + 2𝑦
𝑥+𝑦 =7
Example 2: Solve by elimina on
3𝑥 − 4𝑦 = 7
5𝑥 − 6𝑦 = 8
Page | 22
Example 3: Solve by elimina on
2𝑥 + 7𝑦 = 24
3𝑥 − 2𝑦 = −4
Example 4: Solve by elimina on
2
1
𝑥− 𝑦=4
3
2
1
1
5
𝑥− 𝑦=
2
4
2
Page | 23
Lesson 7 – Solving Word Problems by Elimination
Example 1: Gerald has $4.65 in his piggy bank. He has 2 loonies and a collec on of quarters
and dimes. If there are a total of 15 coins in his piggy bank, how many of those are quarters
and how many are dimes?
Example 2: The perimeter of the elephant’s rectangular enclosure is 38 km. If the length is 3
km longer than the width, what are the dimensions of the enclosure?
Page | 24
Example 3: The price of boots is twice as much as a pair of shoes. If two pairs of boots and
three pairs of shoes cost $266.00, how much does each pair of boots and shoes cost?
Example 4: Quinn has two numbers in his head. The sum of the two numbers is 79. The
difference between the first number and 3 mes the second number is 19. What are the two
numbers?
Page | 25
Unit 6: Trigonometry
Quiz:
S.G Quiz:
Unit Test:
Page | 26
Unit 6 Outline: Trigonometry
Homework or practice problems are a necessity to understand and do well in mathematics. If you have
di iculty with your homework, read your notes, read the examples in the textbook, ask questions in class,
and attend extra help session before or after school.
Absent students are responsible for all assigned work, regardless of whether the absence is excused or
not.
You do not have to do both A and B homework where indicated, complete A for the average assignment and B
if you are looking for a more of a challenge.
Day
Topic
Page/Source
Assignment
1
Review Pythagorean Theorem
Worksheet 1
all
p. 75
A: #3ab, 4ac, 5ac, 12, 14
B: #4cd, 5ac, 15, 18, 21
2
Solving for Angles
p.95
p.82
3
A: #6, 7ab, 8ab, 9ab, 10, 12
B: #4, 7, 9, 12
A: #3ab, 5ab, 7, 10
B: #4bc, 5bc, 8, 10, 15
Solving for Sides
p.101
A: #3, 6, 11
B: #4, 7, 9, 12
4
Quiz
Date:
5
Applying Trig. Ratios
p.111
A: # 3, 4, 6ab, 8, 10, 12
B: #3, 5, 6cd, 9, 11, 15
6
Multiple Triangles
p.118-119
A: # 3ac, 5ac, 6,8, 9, 10, 14
B: #3bd, 5bd, 6,8, 9, 14, 19
7
Study Guide Quiz
Due:
8
Unit Exam
Date:
Page | 27
Unit 6 Study Guide: Trigonometry
1. Using the triangles below state the indicated ratios:
a)
b)
∠𝑅 = 90
Find 𝑒 =
R
cos 𝐶 =
tan 𝑇 =
sin 𝐸 =
tan 𝐷 =
21
E
29
D
2. Properly label triangle ∆𝑃𝐴𝑇 𝑤ℎ𝑒𝑟𝑒 ∠𝑃 = 90 𝑎𝑛𝑑 𝑎 ≤ 𝑡.
Label all the angles and all the sides with the correct letters.
3. Determine the indicated variable. SHOW YOUR WORK. For sides, round to nearest tenth. For
angles round to the nearest degree.
a)
c)
e)
b)
d)
f)
Page | 28
4. You are standing 350 ft away from a skyscraper that is 750 ft tall. What is the angle of elevation from
you to the top of the building? Round to the nearest degree.
5. Solve the following triangle where ∆𝑃𝑀𝑁, ∠𝑃 = 90 , 𝑝 = 115.6 𝑐𝑚, 𝑚 = 60.2 𝑐𝑚
Round angles to the nearest degree, and lengths to the nearest tenth.
∠𝑁 =
∠𝑀 =
𝑛=
6. Determine the unknown length, x. Round only your final answer to the nearest tenth.
9.6 cm
7. Determine the measure of the angle indicated, 𝜃. Round your answer to the nearest degree.
A radio tower is built in two sections. From a point 93 feet away, the
angle of elevation to the top of the lower section is 25 o, and the angle
of elevation to the top of the upper section is 40o. How tall is the
upper section of the tower? Round your final answer to the
nearest tenth.
93 feet
Page | 29
Lesson 1 – Pythagorean Theorem
REVIEW
Pythagorean theorem is an equation used to calculate side lengths of a right angle
triangle.
No angles are used to calculate the length of each side using this theorem.
𝑃𝑦𝑡ℎ𝑎𝑔𝑜𝑟𝑒𝑎𝑛 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 =
Uppercase letters represent _______________
Lowercase letters represent _______________
Example 2: Calculating a side. Round to
hundredths of a cm.
7 cm
3 cm
Example 1: Calculating the hypotenuse.
10 cm
4 cm
Page | 30
Give it a try:
1)
2)
7 cm
15 cm
9 cm
Labelling Triangles
Label the following triangles using uppercase letters for ____________, and lowercase
letters for ________________. Notice no angles are referenced.
The following diagram is a right-angle triangle with a reference angle ∠𝐴. Label the
hypotenuse, and the sides opposite and adjacent to ∠𝐴.
A
Page | 31
Relative to ∠𝐵, label the hypotenuse, adjacent, and opposite sides.
B
Using these given sides, we will define the three primary trigonometric ratios.
The sine ratio, the cosine ratio, and the tangent ratio.
Primary Trigonometric Ratios
SOH
CAH
TOA
Use the triangle below to create trigonometric ratios.
B
Cos A =
Cos B =
Sin B =
Tan A =
12
7
A
8
Page | 32
Use a calculator to calculate the value of each trigonometric ratio to four decimal places.
Make sure your calculator is set to degree mode.
sin (32° ) =
tan(71° ) =
(Round to hundredths)
cos (11° ) =
What do these answers mean?
Page | 33
Lesson 2 – Solving for Angles
Sine Ratio
The sine ratio is:
sin(𝐴) =
To calculate angle (A), we take the sin-1 (sin
inverse) of the trigonometric ratios.
A
𝐴 = sin
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
Example 1: Label the sides of the triangle relative to ∠𝑋. Calculate ∠𝑋 and ∠𝑌.
Y
28m
16m
Z
X
Recall: The triangle angle sum theorem states that all angles inside a triangle always
adds up to _________.
Page | 34
Cosine Ratio
The cosine ratio is:
cos(𝐴) =
To calculate angle (A), we take the cos-1 (cos
inverse) of the trigonometric ratios.
A
𝐴 = 𝑐𝑜𝑠
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
Example 2: Label the sides of the triangle relative to ∠𝑌. Calculate ∠𝑌 using cosine.
Calculate ∠𝑋.
Y
23m
11m
Z
X
Give it a try.
Calculate the angles for each triangle below to the nearest degree.
15
8.1
m
X
7
14.5
m
Y
Page | 35
Tangent Ratio
The tangent ratio is:
tan(𝐴) =
To calculate angle (A), we take the tan-1 (tan
inverse) of the trigonometric ratios.
A
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
Example 3: Calculate ∠𝐾 and ∠𝑁 to the nearest tenth of a degree.
𝐴 = 𝑡𝑎𝑛
M
13
K
K=
M=
9
N
Angles of Elevation and Depression
Horizontal
Always start from the horizontal when identifying angles of elevation and depression.
The angle of elevation is measured from the horizontal and up.
While the angle of depression is measured from horizontal and down.
Page | 36
Example 4: An observer is sitting on a dock, watching a float plane. At a certain time, the
plane is 300m above the water and 430m from the observer. Determine the angle of
elevation of the plane measured from the observer to the nearest degree.
Example 5: A bird is perched at the edge of the school roof looking down to the tarmac at a
half-eaten sandwich . The bird is 28.1m high and the sandwich is 200m away from the
wall of the school. What is the angle of depression? Round your answer to the nearest
whole number.
Practice Problem: Determine the angle of elevation of the roof of the house.
Page | 37
Lesson 3 – Solving for Side Lengths
Warm Up
Identify the trigonometric ratios of the following triangle and calculate the values to four
decimal places.
Calculate ∠𝐴 and ∠𝐵 to
sin (A) =
B
the nearest degree.
cos (A) =
1
3
7
tan (B) =
Reference Material
A
9
S OH
sin(𝜃) =
C AH
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
cos(𝜃) =
T OA
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
tan(𝜃) =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
Finding Side Lengths
Use the sine ratio to calculate the unknown lengths of each triangle below. Round to
tenths.
P
a)
b)
Y
38°
q=
32m
9cm
Z
y=
X
R
67°
Q
Page | 38
Use the cosine ratio to calculate the unknown lengths of each triangle below. Round to
tenths.
P
a)
b)
Y
q=
19m
40°
Z
y=
67°
R
X
3.6
m
Q
Use the tangent ratio to calculate the unknown lengths of the triangles. Round to tenths.
a)
Y
X
b)
z=
w=
V
42°
X
5.0
cm
7.2 cm
70°
Z
W
Your turn. Determine which ratio and solve for each unknown side length. Round to
tenths. Y
a)
b)
x=
Z
57°
14.9
cm
43°
X
x=
26 mm
Page | 39
c)
d)
X
V
37 cm
19.25 mm
Y
Z
40°
x=
z=
33°
X
W
Word Problem
The diagram shows measurements taken by surveyors. Calculate the distance between
the transit and the survey pole to the nearest meter.
Page | 40
Lesson 4 – Solving Triangles
What does it mean?
The phrase “solve a triangle” means to calculate all unknown sides and angles of a
triangle.
Example 1: Solve the triangle FGH
F
36°
H
15
G
Give it a try. Solve the following triangle. Round lengths to the nearest tenth and angles to
the nearest degree.
G
9.0 cm
39°
H
J
Page | 41
Example 2: The light on a lighthouse is 25m above the water level. From a position beside
the light, the angle of depression to the sailboat is 12°. How far, to the nearest tenth, is the
sailboat from the lighthouse?
Page | 42
Lesson 5 – Solving with Multiple Triangles
Example 1: Calculate the length of XY to the nearest tenth of a centimeter.
Y
22°
8.4 cm
20°
X
Z
W
Example 2: Calculate all angles in the triangle below.
X
4
cm
7
cm
Z
5 cm
W
Page | 43
Example 3: Find the length of BC.
B
A
32°
12
cm
46°
C
D
Example 4: Calculate angle BEC to the nearest degree.
E
4.6m
3.7m
39°
B
C
A
Page | 44
Example 5: A surveyor stands at a window on the 3rd floor of a building. He uses a
clinometer to measure the angles of elevation and depression of the top and the base of a
taller building. Calculate the height of the taller building to the nearest tenth of a meter.
31°
39m
42°
Example 6: Two wires support a flagpole, FH. The first wire is 11.2m long and has an angle
of elevation of 39°. The second wire has an angle of elevation of 47°. How tall is the
flagpole to the nearest tenth of a meter?
F
G
39°
E
47°
H
Page | 45
35m
Example 7: A communications tower is 35m tall. Due north, Tannis measures the angle of
elevation to the top of the tower to be 70°. Her brother Leif, standing east of the tower,
measures the angle of elevation to the top of the tower to be 50°. How far apart are they,
rounded to the nearest meter?
50°
Leif
70°
Tanni
s
Page | 46
Lesson 6 – Word Problems with Multiple Triangles
Reference Material
The angle of _________ is measured from horizontal upwards, while the angle of _________
is measured from the horizontal downwards.
Example 1: The shadow of a tree is 10m. From the base of the shadow, the angle of
elevation up to the sun is 64°. How tall is the tree, to the nearest tenth of a meter?
Page | 47
Example 2: The distance between two buildings is 500m. A person standing on top of the
shorter building measured the angle of depression to the base of the taller building is 64°.
The angle of elevation from the same location to the top of the tower is 46°. How tall is the
second building, rounded to the nearest tenths?
Example 3: Henry is standing in front of a tree with a tower in the background. The height
of the tree is 11m tall, and the tower is 181m tall. Henry, who stands 1.8m tall, measures
the angle of elevation to the top of the tree to be 33°, and 55° to the top of the tower. How
far apart is the tree from the tower, rounded to the nearest meter?
Page | 48
Unit 7: Measurement
Quiz:
S.G Quiz:
Unit Test:
Page | 49
Unit 7 Outline: Measurment
Homework or practice problems are a necessity to understand and do well in mathematics. If you have
di iculty with your homework, read your notes, read the examples in the textbook, ask questions in class,
and attend extra help session before or after school.
Absent students are responsible for all assigned work, regardless of whether the absence is excused or
not.
You do not have to do both A and B homework where indicated, complete A for the average assignment and B
if you are looking for a more of a challenge.
Day
1
Topic
Imperial Measures of Length
Relating SI and Imperial Units
Page
Assignment
Pg. 11
A: # 3, 4, 7, 10, 13
B: #4, 7, 9, 10, 13, 14
Pg. 22
A: # 4-8, 9, 10, 11, 13
B: #4def, 5cd, 6-9,11, 13, 17
Review: Area of 2D shapes
2
Surface Area of square based
Pyramids and Cones
Pg. 34
3
Surface Area of Pyramids,
Prisms, Cylinder, and Spheres
4
Quiz
5
Volume of Pyramids, cones, and
Spheres
6
Volume and Surface Area of
Composite Shapes
Pg. 59
7
Study Guide Quiz
Due:
8
Unit Exam
Date:
Pg 51
A: # 4, 5, 6, 7, 8, 11
B: #4a, 5, 6a, 7, 11, 18, 21
#3, 5, 7, 8, 9, 11a,
Date:
Pg. 42
# 4, 6, 8, 9, 12, 14, 18
Pg. 51
#4, 11b, 14, 15, 23
A: # 3, 5, 6, 9, 10,
B: #3, 5, 6, 8, 10, 12
Page | 50
Unit 7 Study Guide: Measurement
1. What referent can be used to estimate the following measurements. Place the letter in the box next to the
measurements that best match.
Letter
Measurement
1 yard
1 inch
1 foot
1 mm
1m
1 cm
A
B
C
D
E
F
Width of a pinky finger
Thickness of a dime
Length of an adult male shoe
Width of a thumb across the joint
Height of a doorway
Length of an adult male stride
2. Find the heights of all these students in cm. Order the students from shortest to tallest.
Rory = 1.6 m
Quinn = 64 inches Pat = 184 cm
Sydney = 1.9
Taylor = 5 ft 7 in
yards
Work
Rank the students by height: shortest to tallest.
,
,
,
,
3. Convert the following measurements. Show your work. Round to nearest tenth if necessary.
Original Measurement
Work
Final Answer
31 yds
Ft
12 feet
M
274 cm
In
4yd 2ft 5 in
In
10867 feet
mi
yd.
ft
278 inches
yd
ft.
in
Page | 51
4. Mr. Ha is building a fence for his yard. The dimensions of the yard are 37 feet by 11 yd. How many feet of
fencing will be required? Fencing is sold only by the whole meter and costs $11.75/m. What will be the
cost before tax?
Number of feet required is
.
Total cost before tax is
5. Find the Surface Area (every touchable surface) of the following: (6 marks) Show your work for marks.
Round to the nearest tenth if required. Use π from the calculator.
Square Pyramid ( measurements in cm) Ice Cream Cone – open top
Bowling Ball
14
14
Surface Area =
r = 3 cm
Surface Area =
Surface Area =
6. Find the outer surface area of this pyramid topped prism Round final answer to nearest in 2. (2 marks)
Page | 52
7. A cone has a surface area of 283 m2 and a diameter of 10 m.
a) Sketch the cone and find the slant height rounded to the nearest tenth.
b) What is the internal height rounded to the nearest tenth?
8. The volume of a bouncy ball is 113.1 cm3.
a) Calculate the radius to the nearest tenth
b) Four bouncy balls are stacked on top of each other in a cylinder. How tall is the cylinder to the
nearest tenth?
9. Given a cube with each side equal to 8 cm. What is the volume of the cube?
Sketch the largest cylinder which
will fit inside the cube. What is its
volume?
What is the volume of air which is
inside the cube but not inside the
cylinder?
Sketch the largest sphere which
will fit inside the cube. What is its
volume?
10. Find the volume of each of the following (Round to the nearest tenth)
Cylinder
Cone
Tetrahedron (Triangular
Pyramid)
Sphere
7m
16m
14 m
Volume =
Volume =
Volume =
Volume =
Page | 53
11. Calculate the volume of the following shape. Do not round until the last statement.
marks
Cone Section
3
5 cm
Cylinder Section
8 cm
Hemisphere Section
Total Volume – (Round to the nearest tenth)
Page | 54
Lesson 1 – Imperial Measures of Length
Referent – A common item used to estimate a measurement.
Inch is the most common smallest unit of measurement for imperial units. Compared to
millimeters in the metric system, which is more accurate?
Converting between imperial Units
1. Convert 5 yd to:
a) Feet
b) Inches
2. Convert 51 inch to:
a) Feet and inches
b) Yards, feet, and inches
Page | 55
3. Convert 26,500 ft to:
__________ mi _____________ yds ___________ in
4. Convert 58,850 ft to:
__________ mi _____________ yds ___________ in
5. Convert 3 mi to:
a) ______________________ inches
Page | 56
Problem Solving
Tyrell has 4 yd of chord to make friendship bracelets. Each bracelet needs 8 in of cord.
How many bracelets can he make?
Ben buys baseboards for a square bedroom with a perimeter of 37 ft. What is the length of
baseboards needed in yards and in?
Page | 57
Lesson 2 – Converting between SI and Imperial
Warm Up
Convert 432 inches to …
yards
feet
Convert 1,274 ft to…
_______yards ________feet _________inches
Metric to Metric Conversion
Standard International (SI) Units are metric units of measurement based on a decimal
system, meaning it’s based on powers of 10.
Deca
Choc
Page | 58
Convert 180 cm to…
mm
dm (deci)
Convert 8.53 Km to…
Dm (deca)
cm
Convert 6.44 m to…
hm (hecto)
cm
Imperial ↔ Metric Conversion
Round all answers to the nearest tenth.
Convert 7 ft 4 in to cm
Convert 9m 30cm to yd
Page | 59
Practice Problem #1: A bowling lane is 19 m long. What is the measure to the nearest foot?
Practice Problem #2: A Canadian Football field is 59 m wide. What is the measurement to
the nearest foot?
Practice Problem #3: After meeting in Osoyoos BC, Taryn drove 114 km north and Laurel
drove 68 mi south. Who drove further?
Page | 60
Nora is 5 ft 7 in tall. The height is in cm on Alberta Drivers Licenses. What height will her
license say?
Trucks with loads wider than 4.3 m must have a “wide load” marker and an escort with
flashing lights. Does a 15 ft wide truck meet the requirement? Justify your answer.
Page | 61
Lesson 3 – Review: Pyramids and Cones
2D Area of Shapes
Calculate the perimeter and area of the following 2D shapes, rounding to 1 decimal place.
𝑐 = 𝜋𝑑 = 2𝜋𝑟
𝐴 = 𝜋𝑟
12 cm
6m
5m
4m
P=
𝐴=
6m
2m
P=
𝑏ℎ
2
𝐴=
𝑏ℎ
2
Page | 62
Pyramids
When the base of a right pyramid is a regular polygon, the triangular faces are congruent
(same). The slant height is the height of a triangular face.
A regular polygon is an enclosed shape with equal sides and angles.
Apex
Slant Height
Internal
Height
Regular Tetrahedra
Slant Height
Slant Height
Right-Square Pyramid
Right-Pentagoal Pyramid
The area of the triangular faces of a pyramid is called the lateral area, 𝐴 .
The internal height extends from the apex (tip) to the center of the base.
Anatomy of a Pyramid
Anatomy of a Cone
l = Slant Height
h = Internal Height
s = Base’s side length
Page | 63
Surface area: the amount of space (area) covering the outside of 3D shapes.
Compare the surface area of the Square Based Pyramid below using the formula verses
the sum of all areas.
7 cm
5 cm
5 cm
Formula Method
1
𝑆𝐴 = 𝑠 ∙ 𝑝 + (𝐴
2
Sum of All Sides Method
)
Find the internal height and the SA of the Square Based Pyramid below.
Lateral Edge = 7 cm
6 cm
6 cm
Page | 64
Surface Area of a Cone
The right cone has a base radius of 2 ft and a height of 7 ft. Calculate the surface area to
the nearest square foot.
Page | 65
Lesson 4 – Surface Area of 3D Shapes
Surface Area of Rectangular Pyramids
A right rectangular pyramid has a base with dimensions of 4m by 6m, and a height of 8m.
Determine the surface area to the nearest square meter.
Surface Area of Tetrahedrons
Find the surface area of the regular tetrahedron to the nearest tenth of a square meter.
Surface Area of Cylinders
The following soup can has a diameter of 6cm and a height of 10cm. What is the Surface
Area?
Page | 66
Surface Area of Spheres
The diameter of a baseball is approximately 3 inches. Determine the surfacea area of a
baseball to the nearest square inch. Round to tenths.
A softball has double the diameter of a baseball. How will the surface area be a ected?
Surface Area of Hemispheres
Hemi(Half) + sphere = half a sphere
d = 3.2 Km
Page | 67
Surface Area of Prisms
All prisms are names for their base shape. To find the SA, add the area of each side.
(Drawing a net diagram might help)
What is the name of the following shape?
15 in
121 𝑖𝑛
5 in
What is the surface area of this shape?
Find the SA if the height is 2cm, length is 9cm, and width is 7cm.
Page | 68
Lesson 5 – Volume of 3D Shapes
Comparing volumes of prisms and pyramids with the same base shape
The volume of any prism is 𝑣 = (𝑎𝑟𝑒𝑎 𝑜𝑓 𝑏𝑎𝑠𝑒) ∙ ℎ𝑒𝑖𝑔ℎ𝑡
The volume of any pyramid is 𝑣 =
𝟏
𝟑
(𝑎𝑟𝑒𝑎 𝑜𝑓 𝑏𝑎𝑠𝑒) ∙ ℎ𝑒𝑖𝑔ℎ𝑡
Determine the volume of a right rectangular pyramid with base dimensions 5.4 cm by 3.2
cm, and a height of 8.1 cm. Round to the nearest tenth of a cubic
centimeter.
What is the volume of a right rectangular prism with the same dimensions?
Page | 69
Calculate the volume of a cylinder with a diameter of 12 in and a height of 18 in?
What is the volume of a cone with the same dimensions? Round the nearest whole
number.
Calculating the Volume of a Sphere
The moon approximates a sphere with a diameter of 2160 mi. What is the approximate
volume of the moon?
Determining the Volume and Surface Area of a Hemisphere
A hemisphere has a radius of 5.0 cm. Determine the surface area and volume.
Page | 70
Extra Questions
A cone has a height of 8 m and a volume of 300 𝑚 . Calculate the radius of the base of the
cone to the nearest meter.
The volume of a soccer ball is approximately 113 𝑖𝑛 . What is the diameter of a soccer ball
to the neasrest tenth of an inch?
Page | 71
Lesson 6 – Composite Shapes
Composite Shapes are shapes made up of more than one shape. We need to use and
modify equations of two or more shapes to calculate surface area or volume.
As an example: the composite shape below uses a sphere and a cylinder.
Determine the surface area and volume of this composite object to the nearest tenth.
(Hint: What are the equations used, and how do we modify the equations to represent the shape?)
Page | 72
Calculate the surface area and volume of this composite shape to the nearest whole
number.
SA:
Volume:
Calculate the surface area and volume. Round answers to the nearest tenth of an inch.
Page | 73
Calculate the surface area and volume. Round answers to one decimal place.
Calculate the exposed surface area, and the volume of the cube. Round to one decimal
place.
Page | 74
Study Guide Answers
Unit 5: Systems of Equations
1. a) One solution
b) No solution
2. a) point of intersection (−1,6)
c) Infinitely many solutions
b) substitute (−1,6) in both equations and LHS must be equal to RHS
3. a) (3,4)
b) substitute (3,4) in both equations and LHS must be equal to RHS
4. a) (3,8)
b) substitute (3,8) in both equations and LHS must be equal to RHS
5. a) (9,23)
b) substitute (9,23) in both equations and LHS must be equal to RHS
6. a) Let a = mass of vitamin C in an apple, p = mass of vitamin C in a peach
𝑎 + 2𝑝 = 17
b) 𝑎 = 3𝑚𝑔
2𝑎 + 𝑝 = 13
𝑝 = 7𝑚𝑔
7. a) Let x = the larger number, y = the smaller number
𝑥 − 𝑦 = 23
𝑥 − 1 = 3𝑦 𝑜𝑟 𝑥 = 3𝑦 + 1
8. b) substitute 11 and 34 in both equations and LHS must be equal to RHS
a) Let n = number of nickels, q = number of quarters
0.25𝑞 = 0.05𝑛 = 5.90
b) There are 7 quarters and 83 nickels.
𝑞 + 𝑛 = 90
9.
Number of lbs.
Value of One lb.
Total Value
Peanuts
x
$1.60
1.6x
Raisins
y
$2.80
2.8y
Mixture
24
Equation: 𝑥 + 𝑦 = 24;
$2.05(24)
1.6𝑥 + 2.8𝑦 = 49.2
15 lbs of peanuts and 9 lbs of raisins
Unit 6: Trigonometry
1. a. 𝐶𝑜𝑠𝐶 =
𝑡𝑎𝑛𝑇 =
b. 𝑒 = 20
𝑆𝑖𝑛𝐸 =
t
A
p
𝑡𝑎𝑛𝐷 =
P
a
T
Page | 75
2.
3. a) 48
4. 𝑋 = 65
b) 𝑥 = 13.8
c) 29
d) 𝑥 = 6
< 𝑁 = 59
N
p
m
e) 36
< 𝑀 = 31
f) 𝑥 = 12.9
𝑛 = 98.7𝑐𝑚
5.
P
n
6. 𝑥 = 18.1𝑐𝑚
7. 𝜃 = 46
8. 𝑥 = 32.4𝑓𝑡
M
Unit 6: Measurement
1.
Letter
F
D
C
B
E
A
Measure
ment
1 yard
1 inch
1 foot
1 mm
1m
1 cm
2. Rory, Quinn, Taylor, Sydney, Pat
3.
Original
Measurement
31yds
Final Answer
12ft
3.7m
274cm
107.9in
4yds 2ft 5in
173in
10867ft
2mi 102yds 1ft
278in
7yds 2ft 2in
93ft
4. 140ft required for a cost of $1645.00
5. SA of Square Pyramid: 537.8cm2. SA of Open Top Cone: SA = 122.5cm2. SA of Bowling Ball: 227.0cm2
6. SA = 147in2
7. a) S = 13.0m
b) h = 12.0m
10
m
8. a) r = 3.0cm
b) h = 24.0cm
9. V = 512cm3. V =402.1cm3. V = 109.9cm3. V = 268.1cm3
10. a) Vcylinder = 538.8m3
b) Vcone = 340.3cm3 c) Vtriangular Pyramid = 340.0cm3 d) Vsphere = 2144.7m3
11. Vcone = 64.1408cm3 , Vcylinder = 307.876cm3 , VHemisphere = 89.7972cm3 , Total Volume = 461.8cm3
Page | 76
Math 10 Common – Formula Sheet
Imperial to Imperial
SI to Imperial
Imperial to SI
SI to SI
1 ft = 12 in
1 yd = 3 ft
1 yd = 36 in
1 mi = 1760 yd
1 mi = 5280 ft
1 mm = 0.03937 in
1 cm = 0.3937 in
1 m = 39.3701 in
1 m = 3.2808 ft
1 m = 1.0936 yd
1 km = 0.6214 mi
1 in = 2.54 cm
1 ft = 30.48 cm
1 ft = 0.3048 m
1 yd = 91.44 cm
1 yd = 0.9144 m
1 mi = 1.6093 km
1 cm = 10 mm
1 m = 100 cm
1 km = 1000 m
𝑆𝐴
= 𝑠𝑢𝑚 𝑜𝑓 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑎𝑙𝑙 𝑠𝑖𝑑𝑒𝑠
𝑆𝐴
=
𝑆𝐴
1
𝑠 ∗ 𝑝 + (𝑎𝑟𝑒𝑎 𝑜𝑓 𝑏𝑎𝑠𝑒)
2
𝑉
= (𝑎𝑟𝑒𝑎 𝑜𝑓 𝑏𝑎𝑠𝑒) ∗ ℎ𝑒𝑖𝑔ℎ𝑡
𝑉
= 𝑙𝑤ℎ
1
= (𝑎𝑟𝑒𝑎 𝑜𝑓 𝑏𝑎𝑠𝑒) ∗ ℎ𝑒𝑖𝑔ℎ𝑡
3
𝑉
= 2𝜋𝑟 + 2𝜋𝑟ℎ
𝑉
𝑆𝐴
= 𝜋𝑟 + 𝜋𝑟𝑠
𝑆𝐴
= 4𝜋𝑟
𝑆𝐴
=
𝑉
= 3𝜋𝑟
𝑉
1
𝑙𝑤ℎ
3
= 𝜋𝑟 ℎ
=
1
𝜋𝑟 ℎ
3
𝑉
4
= 𝜋𝑟
3
𝑉
=
2
𝜋𝑟
3
SOH CAH TOA
𝑆𝑖𝑛𝜃 =
𝑂
𝐻
𝑦 = 𝑚𝑥 + 𝑏
𝐶𝑜𝑠𝜃 =
𝐴
𝐻
𝑚=
𝑇𝑎𝑛𝜃 =
𝑦 −𝑦
𝑥 −𝑥
𝑂
𝐴
𝑦 − 𝑦 = 𝑚(𝑥 − 𝑥 )
𝑎 +𝑏 = 𝑐
𝐴𝑥 + 𝐵𝑦 + 𝐶 = 0
Page | 77
Page | 78