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Technology Supported Learning Course
Mathematics C30
Preparation Package
2009 Printing
Mathematics C30
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Mathematics C30
Preparation Package
Introduction
The spectrum of students enrolling in distance education courses is a wide one, ranging
from students currently in the secondary school setting to adults who have been out of the
classroom for five to ten years, or longer. Designing a course to meet the learning needs of
such a diverse group is certainly a challenge. Where does one begin? Some students have
the skills necessary to dive into the course immediately while others require some review
to brush up their skills before beginning the course.
In the past, review material has not been included with course materials. However, with a
growing adult enrolment, the need and the number of requests for review material has
been growing. As a result, this package has been put together. The Mathematics C30
Preparation Package consists of a review of basic algebra and geometry skills. Covered
topics include: rational expressions, factoring, solving equations, terminology in
geometry, trigonometric functions and geometric proofs. The package also contains
examples and solutions to illustrate the concepts being presented. If you have been out of
the secondary system for awhile, or feel you could use a refresher before beginning the
course, then it would be beneficial to read through this package.
Table of Contents
Part A
Algebra Review ..........................................................................
1
1
4
6
8
9
11
15
20
21
22
24
27
27
31
Part B
Geometry Review ......................................................................
37
37
39
40
41
44
49
53
58
61
62
67
74
87
89
94
98
105
Evaluating Powers ...........................................................................
Properties of Exponents ..............................................................
Multiplying and Dividing Rational Expressions ............................
Review of Factoring ..........................................................................
Expanding Polynomials ..............................................................
Factoring Polynomials ................................................................
Factoring Trinomials ..................................................................
Factoring Special Polynomials ...................................................
Solving Quadratic Equations ...........................................................
Factoring .....................................................................................
Completing the Square ...............................................................
Solving Linear Systems ...................................................................
Substitution .................................................................................
Elimination .................................................................................
Lines and Line Segments .................................................................
Finding the Distance Between Two Points ................................
Finding the Midpoint ..................................................................
Parallel Lines ...................................................................................
Angles and Their Terminology ........................................................
Triangles ...........................................................................................
Parallelograms and Their Properties ..............................................
Coterminal Angles ............................................................................
Principle Angles ...............................................................................
Pythagorean Theorem ......................................................................
Trigonometric Ratios and Trigonometric Functions .......................
Reference Angles ..............................................................................
Congruent Triangles ........................................................................
Congruence Postulates ...............................................................
Proofs in Geometry .....................................................................
Definitions and Properties used in Proofs .................................
Proving Triangles Congruent .........................................................
Proving Corresponding Parts of Congruent Triangles are
Congruent ........................................................................................
112
Part A: Algebra Review
Evaluating Powers
The following table will show you the pattern that is created as exponents of a constant
base decrease by one (1).
Power
Evaluation
Following the pattern shown in the table,
rules have been developed for zero and
negative exponents.
2
5
32
2
4
16
2
3
8
2
2
4
Zero Exponents
2
1
2
For any base a, a  0 :
2
0
1
1
1
2
2
2
2
2
1
4
3
1
8
0
a = 1
Negative Exponents
For any base a, a  0 , x  I :
a
x

1
a
x
or
x
a 
1
a
x
Brackets are very important in powers. The exponent applies to all of the terms
inside the brackets.
•
•
•
(5)0  1
(5)0  1
( -245 x 9 y 4 ) 0 = 1
Conversely, if the term is outside the brackets, or if there are no brackets, the
exponent only applies to the term directly in front of it.
•
 7 x 0 =  7 1 =  7
•
•
•
4 ab 0 = 4 a(1) = 4 a
 8 0 = ( 1)( 8 0 ) = ( 1)(1) =  1
 3 2 = ( 1)( 3 2) =  9
Mathematics C30
1
Preparation Package
Example
Evaluate 5 2 .
Solution
Take the reciprocal of the base and make
the exponent positive.
5 2 
1
52

1
25
Evaluate.
Example
0
3
Evaluate 7 2 4 5 .
5 2
Solution
1
43
Take the reciprocals of the bases with
negative exponents. Treat each power
separately.
•
•
1
 25
5
2
Evaluate the zero (0) exponent.
•
7 0  1 (Any variable to the
power of zero (0) is
equal to one (1).)
0
3
7 4
2 5
5 2
0 5
 72 23
5 4
Write the original expression.
Simplify.
(1)( 32 )
(25 )( 64 )
1

50
Evaluate.
Mathematics C30
4 3 

2
Preparation Package
The calculator key stroke pattern for this question is:
clear ( 7 yx 0 x 2 yx 5 ) / ( 5 yx 2 x 4 yx 3 ) enter
1
The display is 0.02. This is equal to
.
50
•
When there is a power with a negative exponent in the numerator, move the power
from the numerator to the denominator and make the exponent positive.
1
i.e.) 3 2  2
3
•
When there is a power with a negative exponent in the denominator, move the
power from the denominator to the numerator and make the exponent positive.
1
 43
i.e.)
3
4
Example
Simplify 6 a 3 b 7 c by expressing with positive exponents.
Solution
Take the reciprocals of the bases
with negative exponents.
1
a3
1
 7
b
•
a 3 
•
b 7
The power with the variable base, c, has a positive exponent already.
Write the original expression.
Simplify.

6 a 3 b 7 c
6c
3
7
a b
Mathematics C30
3
Preparation Package
Properties of Exponents
Let x and y be any number and a, b, and c be any integer.
Product of Powers
To multiply powers with like bases, add the exponents.
a
b
a+ b
y y = y
Quotient of Powers
To divide powers with like bases, subtract the exponents.
a
b
a-b
y y =y
a
y
= ya - b
b
y
or
Power of a Power
To find the power of a power, multiply the exponents.
( y a )b = y a x b
Power of a Product
To find the power of a product, find the power of each factor and multiply.
 xy c  x c y c
Power of a Quotient
To find the power of a quotient, find the power of the numerator and the
power of the denominator by multiplying the exponents in each.
c
x
xc
   c
y
 y
Zero Exponents
For any base a, a  0 :
0
a 1
Negative Exponents
For any base a, a  0 , x  N :
1
1
x
x
or
a = x
a = x
a
a
Mathematics C30
4
Preparation Package
Example
Simplify
 24 a 3 b 2
.
8 ab 2
Solution
Write the original expression.
=
=
 24 a 3 b 2
8 ab 2
( 24  8 )( a 3  1)( b2  2)
=
( -3)( a 2 )(1)
 3a 2
=
=
2 ab  4 a 6 b 2 
2 4 a 16 b12
=
 8 a 7 b3
=
Use the quotient of powers' rule.
Simplify.
Example


Simplify 2 ab   4 a 6 b 2 .
Solution
Write the original expression.
Use the product of powers' rule.
Simplify.
Example
3
  3 .1 x 2 y 4 
 .
Simplify 

5
 2 .24 x z 
Solution
3
Write the original expression.
Use the power of a quotient rule.
Use the power of a power rule.
=
Simplify.
=
Mathematics C30
5
  3 .1 x 2 y 4 


 2 .24 5 z 
x


( 3 .1) 3 ( x 2) 3 ( y 4 ) 3
=
3
(2 .24 ) 3 ( x 5) 3 z 
 29 .79 x 6 y 12
11 .24 x 15 z 3
 2 .65
9 12 3
x y z
Preparation Package
When there is the same base in the numerator and the denominator, keep
the power on the side of the fraction that has the largest exponent
and apply the quotient of a power rule.
•
x6
1
1
 14 6  8
14
x
x
x
•
y 19
 y 19 3  y 16
y3
Multiplying and Dividing Rational Expressions
The same procedures that are used for multiplying and dividing rational numbers are also
used for multiplying and dividing rational expressions.
Multiply
72
27

.
21
12
The easiest way to do this is to factor each of the numbers in the numerators and
denominators, and then eliminate the factors that are common to both the numerator and
the denominator.
=
=
=
=
72
27

21
12
2  2  2  3  3
3  3  3

3  7
2  2  3
2 2 2 3 3 3 3 3

3 7
223
2  3  3  3
7
54
7
Multiplying rational expressions:
a
c
a  c

=
b
d
b  d
Mathematics C30
6
Preparation Package
Example
2
+ 5a + 6
a a + 2 
 2
Multiply the expressions a 3
.
2
a + a
a + 4a + 4
Solution
2
a a + 2 
a + 5a + 6
 2
3
2
a + a
a + 4a + 4
Write the original expressions.
Factor each of the expressions.
=
a + 3 a + 2 
2
a a + 1 
Reduce by eliminating the common factors.
=
a  3 a  2   a a  2 
a  2 a  2 
a 2 a  1 
Simplify.
=
a + 3 
a a + 1 

a a + 2 
a + 2 a + 2 
Example
Multiply the expressions
2
5x  5
x  25
 2
.
2
x  2 x  15
x + 4x  5
Solution
2
5x  5
x  25
 2
2
x  2 x  15
x + 4x  5
Write the original expression.
Factor each of the expressions.
=
( x + 5)( x  5)
5( x  1)

( x  5)( x + 3)
( x + 5)( x  1)
Reduce by eliminating the common factors.
=
 x  5  x  5   5  x  1 
 x  5  x  3   x  5  x  1 
Simplify.
=
5
x+ 3
Mathematics C30
7
Preparation Package
When dividing rational expressions, use the same rule as when dividing rational
numbers.
•
•
Take the reciprocal of the second term.
Then multiply the two terms.
Dividing Rational Expressions:
a
c
a
d
a  d

=

=
b
d
b
c
b  c
Example
2
2
+ 5 b  14
 4b + 4
 b
Simplify b 2
.
5 b2  10 b
b + 8b + 7
Solution
2
2
b + 5 b  14  b  4 b + 4
2
5 b2  10 b
b + 8b + 7
Write the original expressions.
2
+ 5 b  14
5 2  10 b
 2b
= b2
b + 8b + 7
b  4b + 4
Multiply by the reciprocal.
Factor.
=
Eliminate common factors.
=
Simplify.
=
b + 7 b  2   5 bb  2 
b + 7 b + 1  b  2 b  2 
b  7 b  2   5 bb  2 
b  7 b  1  b  2 b  2 
5b
b  7 ,  1, 2
b+ 1
Review of Factoring
The process of determining polynomials whose product equals a given polynomial is called
factoring.
In order to understand factoring well, it is necessary to know how to expand (or multiply)
polynomials and to recognize the fact that expanding and factoring are inverse operations,
the same way that adding and subtracting are inverse operations.
Mathematics C30
8
Preparation Package
The following illustration shows expanding and factoring with numbers.
Expanding (multiplying)
Factoring
3  7  21
3
 7

21
3 and 7 are factors of 21.
Expanding Polynomials
Expanding polynomials may involve one or many of the following multiplication processes.
Multiplying a Polynomial by a Monomial
The distributive property is used to multiply a polynomial by a monomial.
The Distributive Property
The product of a and (b  c) is given by:
a(b  c)  ab  ac or (b  c)a  ba  ca .
The product of a and (b  c) is given by:
a(b  c)  ab  ac or (b  c)a  ba  ca .
Example


Expand  2 x 5 xy  x 2  1 .
Solution

Write the expression.
 2 x 5 xy  x 2  1
Apply the distributive property.
=
Multiply.
Simplify.
=
=
Mathematics C30
9

 2 x(5 xy)  2 x( x 2 )  2 x(1)
 10 x 2 y + 2 x 3  2 x
Preparation Package
Multiplying Two Binomials
The distributive property is used twice to multiply two binomials together.
Distributive Property
( a  b)( c  d )  ( a  b)c  ( a  b)d
 ac  bc  ad  bd
( a  b) is distributed.
c and d are distributed.
The distributive property is often remembered as the FOIL method.
Multiply the:
F
O
I
L
First terms of the binomials.
Outside terms of the binomials.
Inside terms of the binomials.
Last terms of the binomials.
Example
Multiply x  3 m m  x .
Solution
Write the expression.
x  3m m  x 
Apply the FOIL Method.
O
F
x  3m m  x 
I
L
Simplify.
F
O
I
L
 xm  x 2  3 m 2  3 mx
=  2 mx  x 2 + 3 m 2
Mathematics C30
10
Preparation Package
Squaring a Binomial
•
•
This is when a binomial expression is multiplied by itself.
Here is an example of the distributive property. The FOIL method could also be
used.
Example
Expand  y  2 2 .
Solution
 y  2 2
Write the expression.
Expand.
=
 y  2  y  2
Use the distributive property.
=
y y  2   2 y  2 
Simplify.
=
2
y 2y 2y 4
=
2
y 4y4
When a binomial is squared, a pattern shows how the answer can be expressed in
a simple form.
(a + b)2
=
•
•
•
a2 + 2ab + b2
Square the first term.
Square the last term.
The middle term is 2 times the first and last terms.
Factoring Polynomials
Remember that factoring polynomials is the opposite of multiplying or expanding
polynomials.
There are two common methods for factoring polynomials: greatest common factor and
difference of squares
Mathematics C30
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Preparation Package
Greatest Common Factor
Polynomials are factored using the reverse of the distributive property. Whenever you are
asked to factor a polynomial, the first question should always be:
"Is there a common factor in all of the terms?".
The greatest common factor (GCF) of two numbers can be found by factoring each of the
numbers, finding all the factors that are common to both and multiplying these common
factors together.
The concept can be shown first with numbers.
The greatest common factor of 18 and 42 can be determined by factoring each number
separately.
18 = 2 × 3 × 3
42 = 2 × 3 × 7
The greatest common factor of 18 and 42 is 2  3 = 6 .
The same is true when finding the greatest common factor of algebraic terms in a
polynomial.
Example
Factor 15 x 2 y  30 xy  35 xy 2 .
Solution
Factor each algebraic term.
15 x 2 y = 3  5  x  x  y
30 xy = 2  3  5  x  y
35 xy 2 = 5  7  x  y  y
Determine the common factors.
•
•
•
The common factors are 5, x and y.
Multiply these common factors together.
The GCF is 5xy.
Write the original expression.
15 x 2 y  30 xy  35 xy 2
Factor the common factor 5xy from
each term.
=
5 xy3 x   5 xy6   5 xy(7 y)
Use the distributive property.
=
5 xy3 x  6  7 y 
Mathematics C30
12
Preparation Package
Check by expanding.
=
5 xy3 x  6  7 y 
15 x 2 y  30 xy  35 xy 2
()
Example
Factor 30 m 2  40 m 2 n  50 m 2 n 2 .
Solution
Determine the GCF of the terms.
•
The GCF is 10 m 2 .
Write the original expression.
30 m 2  40 m 2 n  50 m 2 n 2

Factor the common factor 10m 2 from
each term.
=
10 m 2 3   10 m 2 4 n   10 m 2 5 n 2
Use the distributive property.
=
10 m 2 3  4 n  5 n 2



Again, check the answer by expanding.
Difference of Squares
A difference of squares may occur when you are asked to factor a polynomial that is a
binomial.
There is a special way of factoring an expression where each of the two terms is a perfect
square and there is a negative sign between the two terms. When this pattern occurs it is
called factoring a difference of squares.
Some examples of perfect squares are:
25
a2
16b2c2
5×5
a×a
4bc × 4bc
Some examples of binomials which are a difference of perfect squares are:
x
2
Mathematics C30
 9
y
2
4 a 2  25 b 2
 81
13
Preparation Package
To factor a difference of squares, follow these steps:
•
Find the principal (positive) square root of the first term.
•
Find the principal (positive) square root of the last term.
•
Multiply the sum of these two square roots by the difference of these two square
roots.
Difference of Squares
In general:
2
2
a  b = a + b a  b
Example
Factor y 2  81 .
Solution

Find the principle square root of the first term.
2
y =y

Find the principle square root of the last term.
81 = 9
Write the original expression.
y
Write the product of the sum
and difference of the roots.
=
2
 81
(y+9
)(y–9
)
su
m
o
fth
e
sq
u
a
rero
o
ts
d
iffe
re
n
c
eo
fth
e
sq
u
a
rero
o
ts
Example
Factor 64 a 2  9 b 2 .
Solution
•
•
The square root of the first term is 8a.
The square root of the last term is 3b.
Write the original expression.
Write in factored form.
Mathematics C30
64 a 2  9 b 2
8 a +
=
14
3b8 a  3b
Preparation Package
8 a +
Check by expanding.
3b8 a  3b
=
64 a 2  24 ab + 24 ab  9 b 2
=
64 a 2  9 b 2 ()
Factoring Trinomials
Trinomials can be factored by using a specific pattern.
When the coefficient of the x2 term is equal to 1, use the following pattern.
•
Find two numbers such that:
•
their sum is the coefficient of the 2nd term.
•
their product is the last term.
Example
Factor x 2  6 x  8 .
Solution
Find two numbers whose sum is 6 and whose product is 8.
•
The product of 2 and 4 is 8 (the last term of the trinomial).
•
The sum of 2 and 4 is 6 (the coefficient of the middle term of the trinomial).
Factor.
sum
x 2  6 x  8   x  4  x  2 
product
Example
Factor y 2 + 13 y + 30 .
Solution
Find two numbers whose product is 30 and whose sum is 13.
•
The two numbers are 3 and 10.
Mathematics C30
15
Preparation Package
Write the original expression.
Factor.
=
2
y + 13 y + 30
 y + 3  y + 10 
Trinomials can also be in the form x 2  bx  c where the middle term is negative. When
looking for factors, use the same pattern of factoring the product, or last term, but
remember that when the middle term is negative and the last term is positive, the two
factors will both be negative.
Example
Factor g 2  21 g + 80 .
Solution
Find two numbers whose product is 80 and whose sum is  21 .
•
The two numbers are  5 and  16 .
Write the original expression.
g
Factor.
=
Check by expanding.
=
=
2
g
g
 21 g + 80
 5  g  16 
 5  g  16 
g  16 g  5 g  80
g 2  21 g  80 ()
2
If the last term of the trinomial is negative, then the factors of the trinomial will have
different signs.
Example
Factor a 2  4 ab  32 b 2 .
Solution
Find two numbers whose product is  32 and whose sum is  4 .
•
The two numbers are 4 and  8 .
Write the original expression.
a 2  4 ab  32 b 2
Remember to factor the b2
in the last term.
Factor.
Mathematics C30
=
16
a  8ba  4 b
Preparation Package
There are many different ways to factor a trinomial where the coefficient of the x2 term is
not equal to 1.
One method of factoring these trinomials is to use trial and error.
•
This can be done by finding the factors of the first term, finding the factors of the
last term and writing down all the possible combinations. Expand each of the
expressions to see which set of factors gives you the correct middle term.
Trinomials of this form can also be factored using decomposition.
•
A pattern is used to factor trinomials where the x2 term is not equal to 1. Factoring
by grouping is one of the steps.
The steps are:
•
Multiply the coefficients of the first and last terms.
•
Find two numbers whose product is this number and whose sum is the coefficient of
the middle term.
•
These two numbers now become the middle two terms.
•
Group the first two terms together and the last two terms together.
•
Remove a common factor from each.
•
Factor by grouping.
Example
Factor 15 y 2  7 y  4 by the trial and error method.
Solution
•
Pairs of factors of 15 are:
(15, 1)
(1, 15)
(5, 3)
(3, 5)
Mathematics C30
•
17
Pairs of factors of  4 are:
(4 , 1)
(1,  4 )
(4,  1 )
(  1 , 4)
(2,  2 )
(  2 , 2)
Preparation Package
Create a table of possible solutions.
Middle term
11y
 59 y
 11 y
59y
28y
 28 y
7 y
 17 y
7y
17y
4y
4y
(15 y  4 )(1 y  1)
(15 y  1)(1 y  4 )
(15 y  4 )(1 y  1)
(15 y  1)(1 y  4 )
(15 y  2)(1 y  2)
(15 y  2)(1 y  2)
(5 y  4 )( 3 y  1)
(5 y  1)( 3 y  4 )
5 y  4 3 y  1 
(5 y  1)( 3 y  4 )
(5 y  2)( 3 y  2)
(5 y  2)( 3 y  2)
correct middle
term
Therefore, 15 y 2  7 y  4  (5 y  4 )( 3 y  1) .
Example
Factor 15 y 2  7 y  4 by decomposition.
Solution
Multiply the first and last terms.
15  ( 4 )  60
Find two numbers whose product is  60 and whose sum is 7 (coefficient of middle term).
•
These two numbers are 12 and  5 .
Rewrite the expression with these two numbers as coefficients for the middle terms.
=
Group terms together and factor.
3 y(5 y  4 )  1(5 y  4 )
=
(3 y  1)( 5 y  4 )
=
Check by expanding.
(3 y  1)( 5 y  4 )
15 y 2  12 y  5 y  4
15 y 2  7 y  4 ()
=
=
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Preparation Package
Example
Factor 6 m 2 + mn  15 n 2 by decomposition.
Solution
Multiply the first and last terms.
6   15  =  90
Find two numbers whose product is  90 and whose sum is 1.
•
These two numbers are  9 and 10 .
Rewrite the expression with these two numbers as coefficients for the middle terms.
6 m 2 + mn  15 n 2
=
Group terms together and factor.
Write in factored form.
=
6 m 2  9 mn + 10 mn  15 n 2
=
3 m (2 m  3 n )  5 n (2 m  3 n )
3m +
5n 2m  3n 
The greatest common factor can also be determined from mathematical expressions. The
greatest common factor of two expressions can be found by factoring each of the
expressions, finding all the factors that are common to both and multiplying these
common factors together.
Example
Find the greatest common factor from the expressions 6 c 3  6 c 2  36 c and 3 c2  27 c  66 .
Solution
Factor each expression. See if you can get the listed factors for each expression by using
the decomposition method!
6 c 3  6 c 2  36 c  6 c ( c 2  c  6 )
 3  2  c  ( c  2)  ( c  3)
3 c 2  27 c  66  3  ( c  11 )  ( c  2)
The greatest common factor of 6 c 3  6 c 2  36 c and 3 c2  27 c  66 is 3  ( c  2)  3( c  2) .
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Preparation Package
Factoring Special Polynomials
Difference of Squares of Special Polynomials
The two processes of grouping and factoring a difference of squares will be incorporated
into this section.
When grouping within a polynomial, pairs of terms were grouped together.
It is also possible to group together three terms that form a special polynomial. This
special polynomial will usually be a trinomial square. It is therefore necessary to be able
to recognize a perfect trinomial square at a glance. This is the first step.
The second step involves factoring a difference of squares.
Example
Factor x 2  4 x  4  25 .
Solution
Write the original expression.
x 2  4 x  4  25
Group together the terms that form
a special polynomial.
=
x
2

 4 x  4  25
Perfect Trinomial Square
Factor the perfect trinomial square.
=
 x + 2 2  25
=
=
( x + 2) + 5 ( x + 2)  5 
x + 2 + 5 x + 2  5 
=
( x  7)( x  3)
Factor as a difference of squares.
Simplify.
Example
Factor x 2  6 x  9   x  4 2 .
Solution
x 2  6 x  9   x  4 2
Write the original expression.
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Preparation Package
Group together terms that form
a special polynomial.
=
Factor the perfect trinomial square.
=
x
2

 6 x  9   x  4 2
Perfect Trinomial Square
Factor as a difference of squares.
 x  3 2   x  4 2
=
( x  3)  ( x  4)( x  3)  ( x  4)
Simplify.
=
=
=
x  3  x  4 x  3  x  4 
2 x  7 (1)
2x 7
Mental Math Calculations
•
Multiply 33  27 mentally.
•
Think of the question as:
30  3 30  3
•
Hint: Square the first terms, square the last terms and
subtract.
•
Now try:
42  58
55  65
26  34
71  69
Solving Quadratic Equations by Factoring
Two basic definitions or rules are important as you work through this section.
A quadratic equation is an equation that can be written in the
standard form
ax 2 + bx + c = 0 where a  0 .
Zero-Product Property
For all a, b   ,
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Preparation Package
if ab = 0 , then a = 0 or b = 0 or both.
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Preparation Package
Factoring
The following steps are used when solving a quadratic equation by factoring.
•
•
•
•
•
•
Write the equation in standard form.
Factor the equation.
Equate each of the factors to zero as shown in the zero-product property.
Solve each of these equations separately.
Check your answers by substituting the solution back into the original equation.
The value or values of the solution are called the roots of the equation.
This method of solving quadratic equations is only useful when the quadratic expression is
easily factored.
Example
Solve the quadratic equation, x 2  6 x  8  0 .
Solution
Write the equation.
x2  6x + 8 = 0
x
 2 x  4  = 0
x  2= 0
x  4= 0
x= 2
x4
Factor the equation.
Equate each of the factors to zero.
Solve each equation.
Check the solutions.
Substitute x = 2 .
x2  6x + 8 = 0
2 2
 6 2  + 8 =? 0
4  12 + 8 =? 0
0 = 0 ()
Substitute x = 4 .
x2  6x + 8 = 0
 6 4  + 8 =? 0
16  24 + 8 =? 0
0 = 0 ()
4 2
The roots of the equation are x = 2, 4 .
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Preparation Package
Example
Solve the quadratic equation 2 x 2  5 x = 3 .
Solution
Write the equation.
Write the equation in standard form.
Factor the equation.
Equate each of the factors to zero.
2x2  5x = 3
2x2  5x  3 = 0
2 x  1x  3  = 0
2x  1 = 0
x  3= 0
1
x3
x=  ,
2
Solve each equation.
Check the solutions.
1
2
Substitute x =  .
2x2  5x = 3
2
 1
 1
2     5    =?
 2
 2
2 5 ?
+ =
4 2
6 ?
=
2
3=
3
3
3
3
2x2  5x = 3
Substitute x = 3 .
2 3 2  5 3  =? 3
18  15 =? 3
3= 3
The roots of this equation are x = 
()
()
1
, 3.
2
This quadratic equation also has the corresponding function.
•
2 x 2  5 x  3 = 0 corresponds to f  x  = 2 x 2  5 x  3
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Preparation Package
Completing the Square
Many quadratic equations cannot be easily solved by factoring. This most likely occurs
when the roots are not integers.
Another way of solving quadratic equations is by completing the square. All quadratic
equations can be solved by completing the square.
A perfect trinomial square is in the form:
a
2
 b  = a 2  2 ab  b 2
or
2
2
a  b  = a  2 ab  b 2
The method of completing the square involves changing a portion of the equation into the
form of a perfect trinomial square.
Another concept that will be used in solving quadratic equations by completing the square
is that of finding the square root of both sides of an equation. It is important to remember
the following rule.
The solution to:
a 2  b is a   b
Example:
1)
a2  4
2)
 x  2 2
7
x 2   7
a  4
a  2
x2 7
Example
Solve the quadratic equation, x 2  2 x  5 = 0 , by completing the square.
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Preparation Package
Solution
•
An attempt to factor this equation does not work because it is difficult to find two
numbers whose product is  5 and whose sum is 2.
Write the equation.
Isolate the constant term.
Find a number to make the left side of the equation
a perfect trinomial square.
•
x2  2x  5 = 0
x2  2x = 5
Take half the "x coefficient" and square that number.
x2 + 2x
1

2
  2  1; 1  1 
2

x2 + 2x = 5
Add this number to both sides of the equation.
x2 + 2x + 1 = 5 + 1
 x + 1 2
Simplify. ( x 2  2 x  1 is a perfect trinomial square.)
Remember, a  b means a   b .
Solve for x.
= 6
x+ 1=  6
x = 1 
2
The two roots of the equation are x =  1 
6 and x =  1 
6
6.
These are the exact roots of this quadratic equation. The approximate values of the roots
to two decimal places can be found by using a scientific calculator.
Use the following key stroke pattern to find the approximate roots of
x = 1  6 :
For x =  1 + 6 :
CLEAR (-) 1 +
6 ENTER
DISPLAY: 1.45
For x =  1  6 :
CLEAR (-) 1 –
6 ENTER
DISPLAY:  3.45
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Preparation Package
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Preparation Package
Example
Solve the quadratic equation, x 2  4 x  7  0 by completing the square.
Solution
Write the equation.
Isolate the constant term.
Find a number to complete the square.
Take half the coefficient of x and
square it
Add 4 to both sides.
x2  4 x 7  0
x2  4 x = 7
2
1

  4   4
2

x2  4 x  4 = 7  4
x
Simplify. ( x 2  4 x  4 is a perfect trinomial square.)
2
 2  = 11
Remember that a 2  b means a   b .
Solve for x.
The two roots of the equation are x = 2 +
x  2 =  11
x = 2  11
11 and x = 2 
11 .
•
If you evaluate for x = 2 + 11 and x = 2 
values will be x = 5.3166 and x =.  1.3166 .
•
One of the strengths of the "completing the square" method of solving quadratic
equations is that the answer can be left as a radical and therefore the roots are
exact.
11 , using a scientific calculator, the
When using the "completing the square" method, the first step is to make sure the
coefficient of the x 2 term is equal to one (1). This can be accomplished by dividing each
term in the quadratic equation by the coefficient.
Example
Solve the quadratic equation, 4 x 2  2 x  1 = 0 by completing the square.
Solution
Write the equation.
4 x2  2x 1 = 0
Coefficient of x 2 term is 4.
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Preparation Package
Divide each term by 4.
2
x 
Mathematics C30
29
1
1
x = 0
2
4
Preparation Package
Isolate the constant term.
x
2

1
1
x=
2
4
Complete the square.
2
 1  1 
1
1
       
16
4
 2  2 
Take half the coefficient of x
and square it.
Add
1
to both sides.
16
x
2

2
1
1
1
1
x+
=
+
2
16
4
16
2
1
5

x   =
4
16

Simplify.
Remember that a 2  b means a   b .
x
Simplify.
x 
1
= 
4
1
= 
4
5
5
16
16
1
5

4
4
1  5
x=
4
Solve for x.
x=
The two roots of the equation are x =
1  5
1  5
and x =
.
4
4
Solving Linear Systems
Substitution Method in Four Steps
1.
Select an equation and isolate one variable (express one variable in
terms of the other).
2.
In the other equation, substitute the expression for the variable and
solve.
3.
Substitute the numerical value into one of the original equations to
solve for the last unknown.
4.
Check to ensure the ordered pair, intersection of the two lines, satisfies
the two equations.
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Preparation Package
Example
Use the substitution method to solve the system.
x= 3
1
Equation one
2 y  3x = 6
2
Equation two
Solution
Step one
(Select an equation and isolate one variable.)
x= 3
Select equation one.
x is already isolated.
1
Step two
(Substitute the expression from Step 1 into the next equation.)
2 y  3x = 6
2
2 y  33  = 6
2y 9 = 6
2 y = 15
15
y=
3
2
Step three
(Solve for the other unknown.)
Write the other equation.
Substitute expression for x.
Solve for y.
Write an original equation.
Substitute value for y.
Substitute 3
into
2 .
Solve for x.
2 y  3x = 6
2
2   3 x  6
 15 
2   3 x = 6
 2 
15  3 x = 6
 3x =  9
x= 3
Step four
(Check the solution.)
 15 
Check  3 ,  .
 2 
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Preparation Package
Write the equations.
Substitute.
2 y  3x = 6
 15 
2    3 3  = 6
 2 
15  9 = 6
6= 6
1
x= 3
(3 ) = 3
2
 15 
Lines x = 3 and 2 y  3 x = 6 intersect at  3 ,  .
 2 
The use of brackets for substitution is a good practice for decreasing the amount of
working errors.
Example
Use the method of substitution to solve the system of linear equations.
x  y= 2
1
x 2y = 7
2
Solution
Step one
Write one equation.
Isolate x. (Express x in terms of y.)
x  y= 2
x = 2 y
Step two
Write the other equation.
Substitute expression from Step 1 for x.
Substitute 3 into 2 .
Solve for y.
Mathematics C30
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3
x + 2y = 7 2

 + 2y = 7
 2 + y  + 2 y = 7
2 + 3y = 7
3y = 9
y= 3
4
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Preparation Package
Step three
Substitute y = 3 into one of the original equations.
Write the equation.
Substitute 4 into 1 .
x  y= 2
x  3  =  2
x= 1
1
Step four
Check 1, 3  .
x  y= 2
1  3  =  2
2 = 2
1
x + 2y = 7
1 + 23 = 7
7= 7
2
Therefore, 1, 3  satisfies both equations.
Example
Solve for x and y using the substitution method.
1
x  y = 1
4
3
y x= 4
4
Solution
1
2
Step one
3
x= 4
4
3
y=  x+ 4
4
Select 2 .
y+
Solve for y.
2
3
Step two
Write the other equation.
x
3
into
x
1 .
Distribute.
Solve for x.
Mathematics C30
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1
  = 1
4
1 3

x   x + 4  = 1
4 4

3
x
x 1 = 1
16
19
x= 0
16
x= 0
Substitute expression for y.
Substitute
1
y = 1
4
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Preparation Package
Step three
Select an original equation.
y+
Substitute 4 into 2 .
y+
3
x= 4
4
2
3
0  = 4
4
y= 4
Step four
Check 0, 4 .
1
y = 1
4
0   1 4  =  1
4
1 = 1
x
1
y+
4  +
3
x= 4
4
2
3
0  = 4
2
4= 4
The solution is 0, 4 .
•
As you have seen, there are three situations that you will encounter when working
with linear equations. Their graphs have one intersection point, no intersection
points, and infinite number of intersection points.
•
When solving a system of equations algebraically, and you do not find a solution,
remember that the lines may be parallel, the same line, or an error has occurred in
your work. A graph of the system will confirm this idea.
Elimination
Systems of equations that we have seen so far have usually had at least one variable
having a coefficient of 1 or  1 .
•
Solutions can still be found using the substitution method on systems having
coefficients other that 1 or  1 , except that it is more likely that the coefficients will
be fractions.
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Preparation Package
Example
Solve using the substitution method.
 3x  5 y =  4
2x  4 y = 2
1
2
Solution
Write the equation.
Solve for x.
Sub 3 into 2 .
 3x  5 y =  4
 3x = 5 y  4
5
4
x = y
3
3
4
 5
2  y +  + 4 y = 2
3
 3
10
8

y+
+ 4y= 2
3
3
10
12
6
8

y+
y=

3
3
3
3
32   23
 y =  
2 3   32
y = 1
(3,  1)
3
4
2 x  4  1 = 2
2x  4 = 2
2x = 6
x= 3
Sub 4 into 2 .
Check
1
 3 x  5 y  4
 3(3)  5( 1)  4
 9  5  4
1
 4  4 
2x  4 y  2
2 ( 3 )  4 ( 1 )  2
6 4  2
2
22 
 3 ,  1 is the intersection point.
This last example is one of the reasons the elimination method was developed - to reduce
areas where errors may occur because of fractions.
In this section we will use the method of elimination to solve systems of equations that
have coefficients of any value.
In order to use this method of solving systems of equations the idea of equivalent
equations must be understood.
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Preparation Package
The Elimination Method in Five Steps
1.
Line up the same variables of the two equations in columns.
2.
Choose a variable to eliminate.
• If necessary, multiply one or both equations by a number so
that the variable chosen to be eliminated has coefficients
which are opposites.
3.
Add the two equations and solve for the other variable.
4.
Substitute the value into one of the original equations to solve for the
eliminated variable.
5.
Check to ensure the ordered pair satisfies the two equations.
Example
Solve the linear system using the elimination method.
y  3x = 8
2x  y = 7
1
2
Solution
Step one
Line up the same variables in columns.
Step two
3x  y = 8
2x  y = 7
1
2
(Eliminate a variable)
Since the coefficients of y are opposites,
adding the two equations will
eliminate the variable y.
3x + y = 8
2x  y = 7
1
2
3x + y = 8
2x  y = 7
5x
= 15
x= 3
1
2
Step three
Add.
+
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Preparation Package
Step four
y  3x  8
(Solve for the eliminated variable.)
1
y  3 3   8
y  1
Write any one of the equations.
Sub 3 into 1 .
Step five
Check 3 ,  1 .
Write the equations.
2x  y = 7
23    1 = 7
7= 7
1
y + 3x = 8
-1 + 33  = 8
8= 8
2
Example
Use the method of elimination on the following system.
x 2y = 0
x y= 3
1
2
Solution
Step one
Line up the variables.
x 2y = 0
x y= 3
1
2
Step two
Eliminate variable x.
* Multiply 2 by  1 . x  y = 3
1
3
x + 2y = 0
 x + y= 3
x can now be
eliminated by
adding
*
We know we can do this because we know if we multiply both
sides of the equation by the same value, the graph of the
equation is unchanged.
Step three
Add the two equations.
1
+ 3
+
Solve for y.
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x + 2y = 0
 x + y = 3
3y = 3
y = 1
1
3
4
Preparation Package
Step four
Write one of the original equations.
Sub 4 into 2 .
x y3
x   1  3
x 1  3
x2
2
Step five
Check 2 ,  1 .
Write the equations.
x 2y =
2   2 1 =
22 =
0=
0
0
0
0
x y= 3
2   1 = 3
2 1 = 3
3= 3
1
2
Example
Find the coordinates of the intersection of the linear system.
1
3
y+
y = 21
2
4
2
1
x 
y= 4
3
6
1
2
Solution
Step one
Eliminate fractions.
Write the equations.
Mult by the LCD.
Distribute.
1
3
y+
y = 21
2
4
3 
1
4 y +
x = 21 4
4 
2
2 y + 3 x = 84
3
Line up 3 and 4 variables.
Mathematics C30
2
1
x 
y= 4
3
6
1
1 
2
6 x 
y = 4 6
6 
3
4 x  y = 24 4
3 x + 2 y = 84
4 x  y = 24
38
2
3
4
Preparation Package
Step two
Eliminate variable y.
Write equation 3 .
Mult 4 by 2 .
4 x  y = 24
Step three
Add 3 and 5 .
+
Solve for x.
3 x + 2 y = 84
8 x  2 y = 48
×2
4
3 x + 2 y = 84
8 x  2 y = 48
11 x
= 132
x = 12
3
5
3
5
6
Step four
1
3
y+
y = 21
2
4
Write the equation.
Sub 6 into 1 .
1
1
3
12  = 21
y+
2
4
1
y + 9 = 21
2
1
y = 12
2
y = 24
Step five
Check 12 , 24  .
Write the equations.
Mathematics C30
1
3
y+
y = 21 1
2
4
1
24  + 3 12  = 21
2
4
12 + 9 = 21
21 = 21
39
2
1
x 
y= 4
3
6
2
12   1 24  = 4
3
6
8  4= 4
4= 4
2
Preparation Package
Part B: Geometry Review
Lines and Line Segments
A straight line, or simply a line is an infinite set of points. It extends indefinitely
in two directions.
This line is called:
•
line m
or
line AB
or

AB
A ray is part of a line that starts at a point and extends indefinitely in one
direction.
This ray is called:
•
ray CD
or
CD where C is the endpoint of the ray.
A line segment, or simply a segment is part of a line that starts at one point and
ends at another point. It includes the two end points and all the points in between.
This segment is called:
•
segment EF
or
EF
The length of the segment from E to F can be written EF. Note that EF is a
number, and EF is a set of points.
Congruent segments are segments that have the same measure.
AB = CD
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Preparation Package
The midpoint of a segment is the point on the segment that divides the segment
into two congruent segments.
MN = NO
A bisector of a segment is any segment, ray or line that intersects a segment at its
midpoint.
CD = DE
Two lines that intersect to form four right angles are perpendicular lines.
•
Line l is perpendicular to line m •
In symbolic terms, l  m .
A perpendicular bisector of a segment is a line that is perpendicular to a given segment
and passes through the midpoint of that segment.
•
m  AB
Mathematics C30
•
AC = CB
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Preparation Package
Finding the Distance between Two Points
The distance between any two points,  x1 , y1  and x 2 , y 2 
can be found by the formula:
Distance =
Example
x 2  x1 2   y2  y1 2
Find the distance between (3, 0 ) and (8 ,  4 ) .
Solution:
x1 , y1
x 2 , y2
 3, 0  and 8 ,  4 
Assign coordinates to the points.
Write the equation.
D
x 2  x1 2   y2  y1 2
8   3 2   4   (0)2
Substitute values.

Simplify.
 (11 ) 2  ( 4 ) 2
 121  16
 137
= 11.8 units
The use of brackets is a good practice!

Would the above example work if the variables are assigned differently?
x 2 , y2
 3, 0 
Mathematics C30
x 1 , y1
8,  4 
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Preparation Package
Finding the Midpoint
The midpoint formula is
 x  x 2 y1  y 2 
M  1
,
 for points  x1 , y1  and  x 2 , y 2 .
2 
 2
Example
Determine the midpoint of the line segment joining point (5, 1) to (13, 7).
Solution
 x1 , y1   x 2 , y2 
5, 1 
13 , 7 
 x  x 2 y1  y 2 
M  1
,

2 
 2
Write the equation.
Substitute known values.
 5  13 1  7 
,


2 
 2
Simplify.
(9, 4)
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Preparation Package
Parallel Lines
Two lines in a plane that do not intersect are parallel lines.
Line t is parallel to line r.
In symbolic terms, t // r .
A transversal is a line that crosses or intersects two or more lines. Transversals that
cross parallel lines have special properties.
Properties of Parallel Lines
Corresponding angles are angles which are in corresponding positions relative
to the parallel lines and the transversal.
•
•
•
l m
a and e are corresponding angles.
They are both on top of a parallel line and to the left of the transversal.
Alternate interior angles are angles on alternate sides of the transversal, and on
the inside of the two parallel lines.
•
•
•
k m
r and s are alternate interior angles.
They are both on the inside of the parallel lines and on opposite sides of the
transversal.
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Preparation Package
The following steps show you why alternate interior angles are congruent:
•
•
•
s and t are vertical angles and they are congruent.
t and r are corresponding angles and they are congruent.
Therefore, r and s , which are alternate interior angles, are also congruent.
Same-side interior angles are angles on the same side of the transversal and in
between the two parallel lines.
•
•
•
a b
k and m are same-side interior angles.
They are on the inside of the parallel lines and on the same side of the
transversal.
The following steps show you why same-side interior angles are supplementary:
•
•
•
k and i are a linear pair and therefore supplementary.
m and i are corresponding angles and therefore congruent.
It follows that k and m which are same-side interior angles, are supplementary.
Example
Given the following diagram with m
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n , solve for a and b.
Preparation Package
Solution:
Solve for a.
1 and 2 form a linear pair and are supplementary.
1 and 2 = 180 
a + 100  = 180 
a = 80 
Solve for b.
2 and 3 are corresponding angles and are congruent.
2 = 3
100  = b  10 
b = 110 
Example
In the diagram m n.
Two same-side interior angles have measures of
4 x + 5 and 20+ x .
Find the measure of each angle.
Solution:
Same-side interior angles are supplementary and the sum of their measures is 180 .
Solve for x.
4 x + 5 + 20 + x =
5 x + 25 =
5x =
x=
Substitute x = 31 for each angle.
4x + 5
= 4(31) + 5
= 129
Check.
129 + 51 =  180
180 = 180 ()
180 
180 
155 
31 
20 + x
= 20 + 31
= 51
Whenever you have parallel lines cut by a transversal, the following statements are true:
•
•
•
•
Corresponding angles are congruent.
Alternate interior angles are congruent.
Same-side interior angles are supplementary.
Angles which form a linear pair are supplementary.
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Preparation Package
Angles and Their Terminology
An angle is formed by two rays with a common endpoint. The common endpoint is
called the vertex and the rays are called the arms of the angle.
This angle is called:
 B
ABC
CBA
When there is only one angle at a vertex it is correct to name the angle with the single
letter which labels the vertex. When there is more than one angle, all three letters must
be used to identify the angle and the letter for the vertex must always be in the middle.
N could refer to any one of the angles MNO , PNO,
MNP. It is therefore necessary to use three letters to
name a particular angle.
Angle
Definition
Acute Angle
an angle with a measure less
than 90°
Right Angle
an angle with a measure of 90°
Obtuse Angle
an angle with a measure
greater than 90° and less than
180°
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Diagram
Preparation Package
Straight Angle
Reflex Angle
an angle with a measure of
180°
an angle with a measure
greater than 180° and less
than 360°
Congruent angles are angles that have the same measure.
mS = mT
Complementary angles are angles whose measures have a sum of 90°.
M and N are complementary
angles
Supplementary angles are angles whose measures have a sum of 180°.
X and Y are supplementary angles.
Adjacent angles share a common vertex and a common side.
BAC and CAD are adjacent angles.
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Preparation Package

A is the common vertex
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AC is the common side
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Preparation Package
Two angles that are both adjacent and supplementary form a straight line.
These two angles form a linear pair.
BAC and CAD are a linear pair.
Example
Two supplementary angles have measures of 2x – 15 and x + 30. Find the
measure of each angle.
Solution:
The sum of the measures of two supplementary angles is 180°.
Write an equation.
1  2  180 
(2 x  15 )  ( x  30 )  180 
Solve for x.
3 x  15  180 
3 x  165 
x  55 
Substitute the value of x to find the measures of the two angles.
1  2 x  15
2  x  30
 2(55 )  15
 95 
 55  30
 85 
A check shows that 95° + 85° = 180°.
The measures of the two angles are 95° and 85°.
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Preparation Package
Vertically opposite angles are the two nonadjacent angles formed when two lines
intersect. They are the angles that are opposite each other.
The angles measuring x° and y° are vertically opposite angles and the angles
measuring w° and z° are vertically opposite angles.
•
mx = my,
•
mw = mz
•
In AOB, O is the vertex, OA is the starting point or initial ray and OB is the
terminal ray. We assume that the angle has been generated by the initial ray OA
rotating about the vertex O until it reaches the position of the terminal ray OB.
•
The angle, in the above figure, is defined as a positive angle because the initial ray
rotated in a counterclockwise direction to reach the position of the terminal ray.
The little arrow near the vertex indicates the direction of rotation.
An angle is in standard position when its initial ray extends horizontally to the right from
the origin. Its initial ray is the positive portion of the x-axis and its vertex is at the origin.
•
An angle in standard position must always be drawn on the co-ordinate axes.
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Preparation Package
Angles are named according to where their terminal arm lies. The x-axis and the y-axis
divide a plane into four quadrants.
If an angle in standard position has its terminal ray in
the first quadrant, it is called an angle in the first
quadrant. If the terminal ray is in the second quadrant,
it is called an angle in the second quadrant, etc.

If an angle in standard position has its terminal ray coinciding
with one of the x or y axis, it is called a quadrantal angle.
Example
Sketch each of the following angles in standard position and state which
quadrant the angle lies in.
1.
25°
2.
15°
3.
225°
4.
325°
5.
180°
Solution:
1.
2.
3.
4.
5.
Quadrant I
Quadrant I
Quadrant III
Quadrant IV
Quadrantal Angle
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Preparation Package
Triangles
A triangle can be classified by the relationship between the sides of the triangle or the
angles of the triangle.
Classification by Sides
Equilateral Triangle
An equilateral triangle
has three congruent
sides.
Isosceles Triangle
An isosceles triangle has
two congruent sides.
Scalene Triangle
A scalene triangle has no
congruent sides.
To show that sides are congruent, a single or double line through the sides is a
symbol that these sides are congruent.
Classification by Angles
Acute Triangle
Equilangular Triangle
Mathematics C30
An acute triangle has
three acute angles.
(Each angle
than 90  )
is
less
An equiangular
triangle is an acute
triangle with all three
angles congruent.
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Preparation Package
Right Triangle
Obtuse Triangle
A right triangle has
exactly one right angle.
An obtuse triangle has
exactly one obtuse
angle
(greater than 90  and
lessthan 180  ).
Example
Classify the following triangles according to both their angles and their sides.
a)
b)
Solution:
a)
According to sides:
According to angles
Equilateral Triangle
Equiangular Triangle
b)
According to sides:
According to angles:
Scalene Triangle
Obtuse Triangle
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Preparation Package
Every triangle can be labeled in the following way.
•
Each of the points, X, Y and Z is a vertex of the triangle XYZ.
•
The side YZ is opposite the vertex X. Sometime YZ is labelled small x, indicating
that it is opposite to  X .
•
Two sides that share a common vertex are called adjacent sides. XY and XZ are
adjacent sides.
In an isosceles triangle, two sides are congruent. These sides
are called the legs of the triangle. The other side is called the
base. The two angles opposite the legs of the triangle are the
base angles of the triangle.
In a right triangle, there is exactly one right angle. The sides adjacent
to this right angle are called the legs of the triangle. The side opposite
the right angle is called the hypotenuse.
The Measures of Angles of a Triangle
The sum of the measures of the angles of a triangle is 180 .
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Preparation Package
Each of the sides of a triangle can be extended to form other angles that are labeled as
exterior angles.
It follows that there is also a unique relationship between the interior angles of a triangle
and the exterior angles.
The measure of an exterior angle of a triangle is equal to the sum of the
measures of the two remote (or non-adjacent) interior angles of the triangle.
Example
Use the figure at the right to answer these questions.
a)
b)
c)
List the interior angles of the triangle.
List the exterior angles of the triangle.
Find the measures of the labeled angles.
Solution:
a)
b)
c)
2 , 5 , and 7
1 , 3 , 4 , 6 , 8 and b
Measures
5 = 75 
Mathematics C30
Reason
5 and a are vertically opposite
angles.
7 = 50 
7 and b are supplementary angles.
2 = 55 
m 5 + m 7 + m 2 = 180 
4 = 105 
m 4 = m 2 + m 7
6 = 105 
m 6 = m 2 + m 7
1 = 125 
m 1 = m 5 + m 7
3 = 125 
m 3 = m 5 + m 7
8 = 130
m 8 = m 2 + m 5
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Preparation Package
Parallelograms and Their Properties
A quadrilateral is a four sided polygon. There are many special
types of quadrilaterals. The first type is a parallelogram. A
parallelogram is a quadrilateral whose opposite sides are parallel.
•
The four vertices of this parallelogram are W, X, Y and Z and therefore it is called
parallelogram WXYZ.
•
The opposite sides of the parallelogram are labeled with arrows showing that the
opposite sides are parallel.
There are special properties that parallelograms have:
If a quadrilateral is a parallelogram, then its opposite sides
are congruent.
Opposite sides of the parallelogram are labeled with lines
showing that the opposite sides are congruent.
If a quadrilateral is a parallelogram, then its opposite angles are congruent.
If a quadrilateral is a parallelogram, then its consecutive angles are supplementary.
 Y +  X = 180
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Preparation Package
Example
From the following diagram, determine the measures of all of the sides and angles. State
the reason for your answer using the properties of a parallelogram.
B = 42
CD = 4
AC = 12
Solution
Angle or Side
Measure
B
42 
A
138 
Consecutive angles of a parallelogram
are supplementary.
D
138 
Opposite angles of a parallelogram are
congruent.
C
42 
Opposite angles of a parallelogram are
congruent.
AC
12
Given
BD
12
Opposite sides of a parallelogram are
congruent.
CD
4
Given
AB
4
Opposite sides of a parallelogram are
congruent.
Mathematics C30
Reason
Given
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Preparation Package
The line that joins the two opposite vertices of a parallelogram is called a diagonal. Each
parallelogram has two diagonals that intersect in one point.
If a quadrilateral is a parallelogram, then its diagonals bisect each other.
Special Parallelograms
A rhombus is a parallelogram that has all four sides congruent.
A rectangle is a parallelogram with all four angles congruent, which
makes each of the angles a right angle.
A square is a parallelogram that is both a rectangle and a
rhombus.
The following Venn diagram shows the relationship among these quadrilaterals.
Remember that in a parallelogram, the opposite sides are parallel.
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Preparation Package
•
Parallel lines have equal slopes.
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Preparation Package
Example
Decide if the following quadrilateral ABCD is a parallelogram by finding the
slopes of its opposite sides.
A (2 , 3) , B (1 , 3) , C (1 ,  1) , D (4 ,  1)
Solution
Plot the points.
Find the slope of AB.
Slope of AB =
3  3
= 0
1  ( 2)
Find the slope of BC.
Slope of BC =
1  3
= 2
1  1
Find the slope of CD.
Slope of CD =
 1  (1)
= 0
 4  (1)
Find the slope of DA.
Slope of DA =
1  3
= 2
 4  ( 2)
The lines are parallel because the opposite sides have the same slopes.
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Preparation Package
Trapezoids
A trapezoid is a quadrilateral that has only one pair of opposite sides that are parallel.
•
The parallel sides are called the bases.
•
The non parallel sides are called the legs.
•
A trapezoid has two pairs of base angles.
The diagonals of an isosceles trapezoid bisect each other.
The midsegment of a trapezoid is parallel to the bases and its length is one half the sum of
the length of the two bases.
Example
Given the following trapezoid, find the measure of the
midsegment.
Solution
•
Let A represent the midpoint of XN and B represent the midpoint of YZ.
•
1
( XY + NZ )
2
1
AB = (4 + 8 )
2
AB = 6
•
•
AB =
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Preparation Package
Coterminal Angles
Three diagrams showing positive angles in standard position.
Previously, standard position was defined and it was noted that the terminal arm of an
angle could fall anywhere in the coordinate system. When the terminal arm is rotated
counterclockwise, as in the above three angles, a positive angle results.
We will now look at negative angles in standard position.
Diagram of three negative angles in standard position.
•
Notice the direction of rotation for negative angles is clockwise.
•
A positive or negative angle is determined by the direction of rotation.
Counterclockwise Rotation = Positive Angle
Clockwise Rotation = Negative Angle
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Preparation Package
Example
In what quadrant does an angle, in standard position lie, if its measure is:
i)
ii)
iii)
 190 
 295 
 160 
Solution
i)
ii)
Quadrant Two
iii)
Quadrant One
Quadrant Three
Different angles, placed in standard position, may have the same terminal arm.
Example:
Angles in standard position whose terminal arms
coincide are called coterminal angles.
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Preparation Package
Example
Determine three positive and three negative coterminal angles for 50  .
Solution
Positive Angles - Counterclockwise Rotation
50   360   410 
50   360   360   770 
50   360   360   360   1130 
Negative Angles - Clockwise Rotation
50   360    310 
50   360    360    670 
50    360    360    360    1030 
There are infinitely many angles coterminal with a given angle.
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Preparation Package
Principal Angles
The principal angle of any given angle in standard
position is the smallest positive angle in standard
position which is coterminal with the given angle.
Example
A)
B)
What is the principal angle whose measure is  240 ?
What is the principal angle whose measure is 840 ?
Solution:
A)
B)
Principal angle
•
120 
Principal angle
120 
If the principal angle has a measure of 50  then all other angles coterminal (as seen
in Example 1) with it can be written 50  + n 360 , n  I , where n is any integer.
I = ...  3 ,  2 ,  1, 0 , 1, 2 , 3...
Check
If n = 0,
the measure is 50  .
If n = 1,
If n = 2,
If n = 3,
the measure is 410  .
the measure is 770  .
the measure is 1130  .
If n =  1 ,
If n =  2 ,
If n =  3 ,
the measure is  310  .
the measure is  670  .
the measure is  1030  .
Principal Angle
}
}
Positive Angles
Negative Angles
General Form - For writing coterminal angles
If A represents any angle, then all angles coterminal with A are represented by
 + n 360 , n  I , where  is the principal angle.
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Preparation Package
Example
Determine the general form of the coterminal angles for an angle whose
measure is  215  .
Solution
Find the principal angle.
•
Sketch given angle.
215° = 180° + 35°
•
Locate smallest, positive angle
180   35   145 
The principal angle measures 145°.
Substitute principal angle into general form equation.
145   n 360 , n  I
Pythagorean Theorem
In any right triangle there are three sides:
the two legs and the hypotenuse.
The hypotenuse is always opposite the right angle.
Pythagorean Theorem
In a right triangle, the sum of the squares of the lengths of the
legs is equal to the square of the length of the hypotenuse.
If a and b are the measures of the legs, and c is the measure of
the hypotenuse then:
a2  b 2  c 2
If you know the lengths of two of the sides of a right triangle, you can find the length of
the third side by using the Pythagorean Theorem.
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Calculator Tip
The square root of a number can be found by using your calculator.
Press the square root key ahead of the number that you want to find
the square root of.
clear
169 enter
display: 13
clear
1587 enter
display: 39.8371685
= 39.84 (rounded to two
decimal places)
Note: You may have a calculator where the
number.
is entered after the
Example
Find the lengths to two decimal places of the missing sides
of this right triangle.
Solution
b)
a = 21
c = 25
a 2  b2  c2
Use the Pythagorean Theorem.
Substitute the known values.
Simplify.
21 2  b 2  25 2
441  b 2  625
b 2 = 184
Take the square root of both sides.
b = 13.5646599  13.56 m
rounded to two decimal places.
clear
Mathematics C30
184 enter
68
Preparation Package
Example
Pete and Cathy want to snowmobile cross country from Wadena to
Kelvington. If they were to drive, they would travel straight north for 22 km
and then go straight east for 25 km. What would be the shortest distance
that they would have to travel if they were to go by snowmobile?
Solution
Read the problem.
The distance from Wadena to Kelvington when you drive is 22 km north and 25 km east.
Find the shortest distance between these two points.
Develop a plan.
Draw a diagram.
Use the Pythagorean Theorem to find the distance of the hypotenuse.
Carry out the plan.
2
2
2
a + b = c
2
2
2
22 + 25 = c
484 + 625 = c2
1109 = c 2
33 .30 = c
clear
109 enter
Check the solution.
2
2
2
?
22 + 25 = 33 .30
1109 = 1109 ()
Write a concluding statement.
The shortest distance from Wadena to Kelvington is approximately 33.30 km.
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What size of screen do you buy when you need to
purchase a computer or a T.V.?
Here are some facts.
•
The size of the screen or monitor is determined by the diagonal of the screen or
monitor.
•
The length and width of a screen or monitor is in the ratio of 4 to 3.
•
The proportion of the length to the width is 4x : 3x.
Example
Computer monitors have different sizes of screens. What is
the difference in the viewable area of a 12 inch screen
compared to a 15 inch screen?
Solution:
Part A: 12 inch screen
Draw a diagram.
3x
4x
Use the Pythagorean Theorem and solve for x.
a 2  b2  c2
(4 x) 2 + (3 x) 2 = 12 2
16 x 2 + 9 x 2 = 144
25 x 2 = 144
x 2 = 5.76
x = 2.4
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Substitute x into 4x and 3x to find the length and width.
4 x  4(2.4 )  9 .6
3 x  3(2.4 )  7 .2
Multiply the length by the width to get the area.
9 .6  7 .2  69 .12 in
2
The viewable area of a computer with a monitor size of 12 inches is 69.12 in 2.
Part B: 15 inch screen
Draw a diagram.
3x
4x
Use the Pythagorean Theorem and solve for x.
2
2
2
a + b = c
(4 x)2 + (3 x)2 = 15 2
16 x 2 + 9 x 2 = 225
25 x 2 = 225
x2 = 9
x= 3
Substitute x into 4x and 3x to find the length and width.
4 x  4(3)  12
3 x  3(3)  9
Multiply the length by the width to get the area.
12  9  108 in
2
The viewable area of a computer with a monitor size of 15 inches is 108 in2.
The difference in viewable areas of the two screens is 38.88 in2.
108 in 2  69 .12 in 2  38 .88 in 2
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Trigonometric Ratios and Trigonometric Functions
The relationships between the angles and the sides of a right-angled triangle are called
the trigonometric ratios or trigonometric functions and these ratios are the foundation of
trigonometry.
In the diagram  is an acute angle in standard position. AB is any perpendicular drawn
to the initial ray OB. Hence,
OA is the hypotenuse,
AB is the side opposite to angle θ, and
OB is the side adjacent to angle θ.
We define the trigonometric ratios of angle θ with reference to the sides of this right-angle
triangle as follows:
sine 
sin  
or
opposite side
AB

hypotenuse
OA
cosine 
or
cos  
adjacent side OB

hypotenuse
OA
tangent 
or
tan  
opposite side
AB

adjacent side OB
cosecant 
or
csc  
hypotenuse
OA

opposite side
AB
secant 
or
sec  
hypotenuse
OA

adjacent side OB
cotangent 
or
cot  
adjacent side OB

opposite side
AB
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Preparation Package
•
Sine, cosine and tangent are often referred to as the three primary ratios in
trigonometry.
*
An easy way to remember the primary trigonometry ratios is to use the
abbreviation:
SOH CAH TOA
Sin  
•
Opp
Hyp
Cos  
Adj
Hyp
Opp
Adj
Notice that the cosecant, secant, and cotangent ratios are reciprocals of the sine,
cosine, and tangent ratios.
csc  =
1
sin 
sec  =
1
cos 
cot  =
1
tan 
Another relationship worth mentioning is tan  =
sin 
=
cos 
=
=
=
•
Tan  
sin 
since
cos 
 AB 


 OA 
 OB 


 OA 
AB OA

OA OB
AB
OB
tan 
Similarly, the relationship cot  =
cos 
can be proven.
sin 
Problems involving the trigonometric ratios can be solved with the use of a scientific
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Preparation Package
calculator. A scientific calculator has the SIN, COS and TAN function keys.
Mathematics C30
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Preparation Package
•
Example: Find sin 37  .
CLEAR SIN 37 ENTER
Display: 0.6018
*
If you did not get 0.6018 as an answer for sin 37  , then you need to
check to see if your calculator is reading 37 rad (radians) instead of
37  .
MODE
With cursor key arrow down  twice and arrow right  once
to have Degree highlighted.
ENTER
CLEAR
(to get back to the original screen)
Try the above box again!
•
Example: Find csc 37  .
CLEAR SIN 37 ENTER
1
ENTER
x
Display:
1.6616
csc  =
Mathematics C30
1
sin 
x
1
=
1
x
1
=
75
1
x
Preparation Package
Example
In the right-angled triangle ABC, AB = 3,
BC = 4, and AC = 5. State the six trigonometric ratios
of C and of A.
Solution
opp 3

hyp
5
adj
4
cos C 

hyp
5
opp 3
tan C 

adj 4
1
4
cot C 

tan C 3
1
5
sec C 

cos C 4
1
5
csc C 

sin C 3
opp 4

hyp
5
adj
3
cos A 

hyp
5
opp 4
tan A 

adj
3
1
3
cot A 

tan A 4
1
5
sec A 

cos A 3
1
5
csc A 

sin A 4
sin C 
Mathematics C30
sin A 
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Preparation Package
The definitions of the basic trigonometric ratios in terms of the sides of a right triangle
can be applied to any acute angle in standard position on the coordinate plane.
With a point P(x,y) on the terminal arm, we can find the distance from P(x,y) to the
origin. This was known as r (as in radius).
•
Start P(x,y) at point S(r,0).
•
As P(x,y) travels about the coordinate system,
circular path is produced.
•
The angle  is the amount of rotation about
the origin.
•
The radius of the circle is the same as the
length of the terminal arm (r).
When finding the distance from P(x,y) to the origin, a perpendicular segment is
constructed from P(x,y) to the x-axis so that a right triangle is produced. With the use of
the Pythagorean Theorem, the distance (r) is found. r 2  x 2  y 2
(Remember, length is always positive.)
The sides of the triangle are x
and y in length and the
hypotenuse has length r.
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Preparation Package
The six trigonometric ratios can now be redefined as follows for any angle in the first
quadrant of measure  in standard position and any point (x, y) on the terminal arm.
sin  =
y
opp
=
r
hyp
csc  =
hyp
r
=
opp
y
cos  =
x
adj
=
r
hyp
sec  =
r
hyp
=
x
adj
tan  =
y
opp
=
x
adj
cot  =
adj
x
=
opp
y
Example
The terminal ray of an angle  in standard position passes through P(8,15).
Evaluate the six trigonometric ratios for  .
Solution
Draw a perpendicular from P to the x-axis to form a right triangle.
2
2
2
r = x  y
2
2
2
r = 8   15 
2
r = 64  225
2
r = 289
r = 17 units
sin  =
y
15
=
r
17
csc  =
r
17
2
=
= 1
y
15
15
cos  =
x
8
=
r
17
sec  =
r
17
1
=
= 2
x
8
8
tan  =
y
15
=
x
8
cot  =
x
8
=
y
15
Using the values for x, y, and r rather than opp, adj, and hyp, has the advantage that
trigonometric ratios may be found of angles in standard position in any quadrant. The
same definitions of the ratios apply. Try the next example when the terminal arm is in
quadrant two.
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Preparation Package
Example
Evaluate the six trigonometric ratios for angle A whose terminal ray passes
through the point  2, 4  .
Solution
2
2
2
r = x + y
2
2
2
r =  2  + 4 
2
r = 4+ 16
2
r = 20
r =
20 = 2 5
x = 2
y= 4
r= 2 5
Note that, although r must always be positive, x and y may be positive or negative
depending on the coordinates of P.
sin A =
=
=
y
4
=
r
2 5
2
csc A =
=
5
r
2 5
=
y
4
5
2
5
5
x
2
=
r
2 5
1
=
5
r
2 5
=
x
2
=  5
cos A =
= 
sec A =
5
5
x
2
=
y
4
1
= 
2
y
4
=
x
2
= 2
cot A =
tan A =
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Preparation Package
Reference Angles
Any angle A in standard position has an associated acute angle called a reference angle.
Reference angles are useful for converting trigonometric functions of any large angle into
a trigonometric function of an angle between 0° and 90°.
The reference angle of a given angle A in standard position
is the positive acute angle determined by the x-axis and
the terminal side of the given angle.
In the following diagrams,  denotes the reference angle of angle A in each of the four
quadrants. ( m A is read 'measure of angle A'.)
In each of the diagrams,  is the positive angle between the terminal arm and the nearest
position of the x-axis.  is always positive and is the reference angle.
QUADRANT TWO
QUADRANT ONE
m  180   mA
m  mA
QUADRANT THREE
QUADRANT FOUR
m  mA  180 
Mathematics C30
m  360   mA
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Preparation Package
Steps to Calculate the Reference Angle
•
Make a sketch of the given angle. The sketch will show if the terminal side of the
given angle is closer to the positive x-axis or to the negative x- axis.
•
Calculate the measure of the angle between the terminal ray and the nearest xaxis. The reference angle is always positive and no greater than 90°.
Example
Sketch each angle whose measure is given and determine the measure of the
reference angle.
a)
b)
c)
m A =  70 
m A = 150 
m A =  210 
Solution
a)
Make sketch of the given angle.
Calculate the measure of the angle between
the terminal ray and the nearest x-axis.
b)
Reference angle has
measure 70°.
Make sketch of the given angle.
Calculate the measure of the angle between
the terminal ray and the nearest x-axis.
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180   150   30 
Reference angle
has measure 30°.
Preparation Package
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Preparation Package
c)
Make sketch of the given angle.
Calculate the measure of the angle between
the terminal ray and the nearest x-axis.
210   180   30 
Reference angle
has
measure 30°.
All coterminal angles have the same reference angles.
Example
Find the reference angle of an angle whose measure is 200° and of an angle
whose measure is  160  .
Solution
Sketch the given angle.
Calculate the measure of the angle between
the terminal ray and the nearest x-axis.
200   180   20 
180   160   20 
Reference angle has measure 20°.
In general form, any angle with measure
200  + n 360 , n  I , has a reference angle of 20°.
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Preparation Package
Why are reference angles important?
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Before this question can be answered, we need to study the outcome of the signs for the
trigonometric ratios depending on which quadrant the terminal arm lies. Since the
coordinates x and y vary in sign from quadrant to quadrant, the trigonometric function
varies in sign from quadrant to quadrant.
Do the following activity to refresh your memory as to which trigonometric function is
positive or negative for an angle whose terminal arm lies in a specific quadrant.
Determine the six trigonometric ratios for each point on the terminal arm of the angle
Fill in the chart.
a) (3, 4)
c)  3,  4 
b) (-3, 4)
d) 3,  4 
Quadrant
One
(3, 4)
Quadrant
Two
Quadrant
Three
Quadrant
Four
r 2  x 2  y2
x= 3
y= 4
r= 5
x = 3
y= 4
r= 5
x = 3
y= 4
r= 5
x= 3
y= 4
r= 5
sin  
y
r
4
5
cos  
x
r
tan  
y
x
csc  
r
y
sec  
r
x
cot  
x
y
 3, 4 

 3,  4 
3,  4 
3
5
4
1
= 1
3
3
Reference
Angle
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Preparation Package
•
Notice that all six trigonometric functions have positive values for angles in the
first quadrant because x, y, and r are each positive.
•
In the second quadrant sine (and its reciprocal ratio, cosecant) is positive but cosine
and tangent (and their reciprocal ratios) have negative values.
•
In the third quadrant tangent (and its reciprocal ratio, cotangent) is positive but
sine and cosine (and their reciprocal ratios) have negative values.
•
In the fourth quadrant cosine (and its reciprocal ratio, secant) is positive but sine
and tangent (and their reciprocal ratios) have negative values.
The above four statements can be summarized into a simple saying that will help you
remember the signs of the trigonometric ratios for any quadrant.
The CAST Rule.
This rule identifies the trigonometric ratios (and the
reciprocals) that are positive in each quadrant.
C - Cosine
A - All
S - Sine
T - Tangent
Example
If A is an angle whose sine is positive, sketch two possible angles in the
correct quadrants.
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Preparation Package
Solution
Since sine is positive in the first and second quadrants the two possible angles are shown
in the diagram. The measure of angle A is not known. The main concern is to draw a
typical angle in the correct quadrant.
Example
Sketch the angle A in the correct quadrant which satisfies simultaneously
the conditions that csc A is negative and cot A is positive.
Solution
For csc A < 0,
angle A is in 3rd
or 4th quadrant.
For cot A > 0,
angle A is in
1st or 3rd
quadrant.
For ccs A < 0, and cot A > 0
together, angle A is in 3rd
quadrant only.
Therefore, an angle in the third quadrant satisfies the conditions.
We are now ready to answer the question that was asked earlier.
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Preparation Package
Why are reference angles important?
If you are using the trigonometric tables to evaluate, you will notice that the tables do not
list angles greater than 90°. The reason is that the trigonometric function of any angle
greater than 90°may always be expressed in terms of the same trigonometric function of
the reference angle.
Example
Express cos 145° in terms of the reference angle and then evaluate.
Solution
Minus signs are used in labelling the sides to remind us that coordinates are
negative.
Sketch the 145° angle and from any point P on the terminal ray
draw a perpendicular to the x-axis.
The reference angle has a measure of 35°.
[By definition of cosine]
x
r
 x
  
r
 cos 35 
 0 .8192
cos 145 
adj
hyp
x
cos 35  
r
cos 35  
use your calculator
or trig. tables
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Preparation Package
 cos 145    cos 35    0.8192
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Preparation Package
Check with your calculator
cos 145 
 cos 35 
cos is negative
in Quad. 2
reference angle
This example shows us …
The value of a trigonometric function of angle A is equal to the value of
•
the same trigonometric function of the reference angle of A with the
appropriate + or – sign (depending on the quadrant of the terminal
side of angle A).
The following example is similar to the example above except that the answer is obtained
quicker when the above principal is applied.
Example
Express cos 145° as a function of the reference angle and evaluate.
Solution
The angle is in the second quadrant and the reference angle is 35°.
cos 145    cos 35 
cosine is negative in
the second
quadrant
  0 .8192
Angles Associated with a Given Ratio
Previously, you worked with a given angle, and using a calculator you were able to find
any of the six trigonometric ratios.
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Preparation Package
Your calculator should be in degree mode for this lesson.
Find:
sin
cos
tan
csc
sec
cot
55  =
103  =
217  =
324  =
117  =
18  =
The inverse function keys on a scientific calculator can find the measure of an angle when
the trigonometric function is given.
The inverse function keys are:
•
•
•
1
sin
1
cos
1
tan
Find .
sin  = 0.8192
cos  =  0.2250
tan  = 0.7536
=
=
=
csc  =  1.7013
sec  =  2.2067
cot  = 3.0777
=
=
=
Using a calculator
The keystroke pattern for finding  for sin  = 0.8192 is:
CLEAR
DISPLAY:
Mathematics C30
2nd sin1 ( 0.8192 )
55.0048
91
ENTER
Preparation Package
Example
Determine all possible values between 0° and 360 for  in each of the following.
a)
sin θ = 0.6363
b)
cos θ = 0.9200
c)
tan θ =  1 .3850
Solution
a)
Sine is positive in the first and second quadrant.
•
The angle set-up will look similar to:
•
sin  = 0.6363
 = 39.52°
Use the following keystroke pattern.
CLEAR
DISPLAY:
2nd sin1 ( 0.6363 )
39.5165
ENTER
 = 39 .52  (to the nearest hundredth degree)
•
Mathematics C30
We now have an angle in the first quadrant ( = 39 .52  ).
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Preparation Package
•
For the angle in the second quadrant, 39.52° becomes the reference
angle.
The values that satisfy sin  = 0.6363 are 39 .52  and 140 .48  .
Check:
(with your calculator)
sin 39 .52  = 0 .6363 ()
sin 140 .48  = 0 .6363 ()
b)
cos  = 0.9200
•
Cosine is positive in quadrants
one and four.
cos  = 0.9200
 = 23.07°
Use the following keystroke pattern.
CLEAR
DISPLAY:
2nd cos 1 ( 0.92 )
23.0739
ENTER
 = 23.07 
•
We now have an angle in the first quadrant ( = 23.07  ).
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Preparation Package
•
For the angle in the fourth quadrant, 23.07  becomes the reference angle.
360   23 .07  = 336 .93 
y
x
The values that satisfy cos  = 0.9200 are 23.07  and 336 .93  .
Check:
(with your calculator)
cos 23 .07 = 0 .9200 ()
cos 336 .93 = 0 .9200 ()
c)
tan   1 .3850
Tangent is negative in quadrants two and four.
tan θ =  1 .385
θ =  54 .17 
Use the following keystroke pattern.
CLEAR
DISPLAY:
2nd tan1 ((–)1.3850 )
 54 .17 
ENTER
 =  54 .17 
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Preparation Package
•
We now have an angle in the fourth quadrant  =  54 .17  .
Reference Angles are always positive acute angles.
•
For the angle in the second quadrant,
 54 .17 = 54 .17  becomes the
reference angle.
180   54 .17  = 125 .83 
The values that satisfy tan  =  1.3850 are  54 .17  (or 305 .83  ) and 125 .83  .
Check:
tan  54 .17  =  1.3850 ()
tan 125 .83  =  1.3850 ()
The calculator will always display one angle and based on the cast rule and
reference angles, you will have to find the solution for the second angle.
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Preparation Package
Congruent Triangles
You will notice that in order to place triangle ABC on triangle XYZ, you must line up
vertices A and X. This means that these two vertices are corresponding vertices. The
angles at these vertices also correspond.
Three pairs of sides and three pairs of angles correspond.
AB  XY
BC  Y Z
AC  X Z
A  X
B  Y
C  Z
The same can be shown by the way the triangles are labelled. The corresponding vertices
are arranged in the same order.
ABC  XYZ
Congruent Triangles
If ABC is congruent to JKL , then the corresponding angles and the
corresponding sides of the two triangles are congruent.
The congruent corresponding parts are marked by a single
line, double lines and triple lines.
Corresponding angles are:
A  J
•
B  K
•
C  L
•
Mathematics C30
Corresponding sides are:
AB  JK
•
BC  KL
•
AC  JL
•
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Example
Name all of the congruent parts of the following two triangles.
Solution
Angles:
•
•
Z  H
A  N
•
Q  B
Sides: •
ZA  HN
AQ  NB
•
•
ZQ  HB
This shows that the definition of congruence is reversible.
If two triangles are congruent,
then the pairs of corresponding angles and
sides are congruent.
If the pairs of corresponding angles
and sides of a triangle are congruent,
Example
then the triangles are congruent.
If FGH  TUV , draw a diagram and mark the diagram to show the
congruent parts.
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Solution
Example
From the following corresponding congruent parts, state the triangles that are congruent.
WB  RF , BY  FP , YW  PR , W  R, B  F, Y  P
Solution
WBY  RFP
Congruence Postulates
Sometimes it is helpful to describe the parts of a triangle in terms of their relative
positions.
•
•
•
•
AB
AB
A
A
is opposite C .
is included between A and B .
is opposite BC .
is included between AB and AC .
Besides the properties or basic terms and definitions associated with geometry, there are
also certain postulates and theorems that are used.
A postulate is a statement that is accepted as being true without proof.
A theorem is a statement that must be proved before it is accepted as being true.
It is not always necessary to have all three sides and all three angles congruent to say
that two triangles are congruent.
The minimum requirements for the proving that two triangles are congruent are stated in
the congruence postulates.
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Side-Side-Side (SSS) Congruence Postulate
If three sides of one triangle are congruent, respectively, to three sides of a second
triangle, then the two triangles are congruent.
Side-Angle-Side (SAS) Congruence Postulate
If two sides and the included angle of one triangle are congruent, respectively, to
two sides and the included angle of a second triangle, then the two triangles are
congruent.
Angle-Side-Angle (ASA) Congruence Postulate
If two angles and the included side of one triangle are congruent, respectively, to
two angles and the included side of a second triangle, then the two triangles are
congruent.
Angle-Angle-Side (AAS) Congruence Postulate
If two angles and the non-included side of one triangle are congruent to two angles
and the non-included side of a second triangle, then the two triangles are congruent.
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Example
For each of the pairs of triangles, state the congruence postulate, if any, that will prove
the triangles congruent.
Solution
a)
b)
ASA Postulate - The congruent side is included between the congruent angles.
SAS Postulate - Two sides and the included angle of one triangle are congruent,
respectively, to the two sides and the included angle of a second triangle.
SSS Postulate - Two sides and the common side are congruent.
None - There is not an AAA Postulate.
c)
d)
Why is there not an AAA Postulate?
Use a protractor and ruler for the following constructions.
Triangle One
•
•
•
•
•
Draw a line 4 cm long. Label the line RS.
Construct an angle of measure 900 at point R.
Construct an angle of measure 300 at point S.
The point where the two lines join will be called T.
You now have RST with angles measuring 30, 60 and 90 degree.
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Triangle Two
•
•
•
•
•
Draw a line 6 cm long. Label the line GH.
Construct an angle of measure 90° at point G.
Construct an angle of measure 30° at point H.
The point where the two lines join will be called I.
You now have GHI with angles measuring 30, 60 and 90 degree.
What can you say about the angles in these two triangles? What can you say
about the two triangles? Is RST  GHI ? (Remember the definition of
congruence triangles)
Example
Each pair of triangles below has two corresponding sides or angles marked as congruent.
Indicate the additional information that is needed to enable you to apply the specified
congruence postulates.
a)
For ASA
b)
For SAS
c)
For SAS
d)
For SSS
e)
For AAS
f)
For ASA
Solution
a)
b)
c)
d)
e)
f)
M  R
NO  ST
C  F
AB  DE
K  Z
T  Y
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An angle is needed on the other side of the included side.
A side is needed on the other side of the included angle.
The angle has to be included between the two sides.
A third side is necessary.
The side is not included between the two angles.
The side is included between the two angles.
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Equivalence relations play a role in understanding the properties of congruences.
The following chart will show you the properties of reflexivity, symmetry and transitivity.
This course will utilize the properties of segment congruence and angle congruence.
Property
Number
Equality
Reflexive
a=a
Symmetric
Transitive
Segment
Congruence
Angle
Congruence
AB  AB
ABC  ABC
If a = b,
then b = a
If AB  CD ,
then CD  AB
If ABC  XYZ
then XYZ  ABC
If a = b and
b = c, then
a=c
If AB  CD and
CD  EF , then
AB  EF
If ABC  XYZ
and XYZ  RST
then ABC  RST
Here are some examples where the reflexive property can be used.
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Statement
Reason
AD  AD
Reflexive Property of
Congruence
X  X
Reflexive Property of
Congruence
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Reflexive Property of
Congruence
CD  CD
Proofs in Geometry
Our minds are often called upon to make decisions. Sometimes these decisions require a
great deal of thought and certain steps must be followed to reach the outcome.
The thought process in these situations is very important.
Activity
Mr. Boyes, Mr. Banadyga, Mr. Crow and Mr. Petersen are gentlemen living in
rural Saskatchewan. They live in Meadow Lake, Kamsack, Estevan and Ponteix.
Use the table and clues to determine where each of the gentlemen live.
a)
Mr. Boyes is a business partner of the men living in southern
Saskatchewan.
b)
Mr. Banadyga is a cousin of the men living in Kamsack and
Estevan.
c)
Mr. Crow and Mr. Petersen just visited northern Saskatchewan.
d)
Mr. Petersen does not live in a city.
Mr. Boyes
Mr. Banadyga
Mr. Crow
Mr. Petersen
Meadow Lake
Kamsack
Ponteix
Estevan
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This has been an example of logical thinking. The most important thing to remember is
that a certain thought process must occur in able for you to solve a problem.
This is the same with proofs in Geometry. Many of the steps in the process of solving
geometric proofs use deductive reasoning.
Deductive reasoning is reasoning that begins with a preliminary statement. Other
statements can be logically developed from this statement through the process of
deduction.
Deductive reasoning is often in the form "If .... then" where the "if" part is the preliminary
statement and the "then" parts are the logical statements that follow from the preliminary
statement.
•
•
•
If it rains out, then the class will not being going on their field trip.
If a person drives a car in Saskatchewan, then he must wear a seat belt.
If a triangle has two congruent sides, then it is isosceles.
Example
Express each of the following statements in the "If ... then" form.
a)
b)
When a  b is a positive number, b is less than a.
A square with a side length of 2 cm has a perimeter of 8 cm.
Solution
a)
b)
If a  b is a positive number, then b is less than a.
If a square has a side length of 2 cm, then it has a perimeter of 8 cm.
Problem solving steps and your ability to use deductive reasoning will help you to prove
that two triangles are congruent.
The problem solving steps are the same ones that you have been using throughout this
course.
•
Read the problem.
•
Develop a plan.
•
Carry out the plan.
•
Look back.
Any time that you are asked to prove that two triangles are congruent, you will be given
information about the two triangles. You will need to take this information and develop a
way to use the information in order to prove the two triangles congruent.
You have learned four ways to prove that two triangles are congruent:
•
SSS Postulate
•
SAS Postulate
•
ASA Postulate
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AAS Postulate
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In each of these cases, you need three congruences of either angles or sides to prove that
the triangles are congruent.
It is important to organize the steps that are necessary in proofs. In this course the
majority of proofs will be done using two columns.
•
The first column will include the statements and show the logical steps that will
result in the proof of the triangles being congruent.
The second column will list the reasons why each statement is true.
•
Example
Given:
ABC and KLM
AB  KL , A  K, B  L
Prove:
ABC  KLM
Solution:
Read the problem.
Given:
ABC and KLM
AB  KL , A  K, B  L
Prove:
ABC  KLM
Develop a plan.
Mark the congruent parts on the two triangles.
AB  KL
(Side)
A  K
B  L
Determine the congruence postulate that can be applied.
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(Angle)
(Angle)
ASA Postulate
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Carry out the plan.
1.
2.
3.
4.
Statement
Reason
AB  KL
A  K
B  L
ABC  KLM
1.
2.
3.
4.
Given
Given
Given
ASA Postulate
Look Back
The final statement of the proof will always be what was originally asked to be proven. In
this example, the last step was that the triangles were in fact congruent, and this was
determined by using the ASA Postulate.
Example
Given:
ABC
AB  AC , BD  CD
Prove:
BAD  CAD
Solution
Read the problem.
Given:
ABC
AB  AC , BD  CD
Prove:
BAD  CAD
Develop a plan.
Mark the congruent parts on the two triangles.
AB  AC
BD  CD
Mark the common side between the two triangles.
AD  AD
Determine the congruence postulate that can be applied.
SSS Postulate
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Carry out the plan.
1.
2.
3.
4.
Statement
Reason
AB  AC
BD  CD
AD  AD
BAD  CAD
1.
2.
3.
4.
Given
Given
Reflexivity property
SSS Postulate
Look back.
The final statement showed that the triangles were proven congruent by the SSS
Postulate. Many times, a common angle or side can be the third pair of corresponding
parts when proving triangles congruent.
Definitions and Properties used in Proofs
Previously, you started proving that two triangles were congruent using a two-column
proof. The congruence postulates that you used were:
•
•
•
•
•
•
Side-Side-Side Postulate (SSS Postulate)
Side-Angle-Side Postulate (SAS Postulate)
Angle-Side-Angle Postulate (ASA Postulate)
Angle-Angle-Side Postulate (AAS Postulate)
Hypotenuse-Leg Theorem (HL Theorem)
Leg-Leg Theorem (LL Theorem)
This section will review a number of definitions and properties that will be used in the
two-column proofs where you will be asked to prove that two triangles are congruent.
Reasons for a statement in a proof will be boxed.
At the beginning of a proof you are always given some information. You may use this
information in the proof by simply stating that the reason is given.
Given
When given information stating that the measures of two segments or two angles are
equal, by definition of congruent segments or angles, it can be deducted that these are also
congruent.
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If d (A,B) = d (C,D) then
If m A = m B
then
AB  CD
A  B
Congruent segments are segments which have the same measure.
Congruent angles are angles which have the same measure.
Recall the three properties stated by the equivalence relations introduced previously: the
reflexive, symmetric and transitive properties. These can be applied to segments, angles,
triangles, or any polygon. Congruence is an equivalence relation on the set of segments
and the set of angles.
A right angle is an angle with a measure of 900.
ABC  XYZ
Any two right angles are congruent.
Two lines that meet or intersect to form right angles are called perpendicular lines.
RS  XY
Perpendicular lines meet to form right angles.
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The midpoint of a segment is the point that divides the segment into two congruent
segments.
C M  DM
Definition of midpoint.
The bisector of a segment is a line, segment, ray or plane that intersects a segment at its
midpoint or bisects the segment into two congruent segments.
RS bisects CD at M
CM  DM
The bisector of an angle bisects an angle into two congruent angles.
YM bisects XYZ
XYM  ZYM
Definition of segment bisector.
Definition of angle bisector.
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The perpendicular bisector of a segment is a line, ray or segment that is:
•
perpendicular to the segment.
•
intersects the segment at its midpoint.
RS  XY
RS bisects XY at M
• RMX , RMY ,
• XM  YM
YMS , SMX are
right angles
• RMX  RMY  YMS  SMX
Definition of a perpendicular bisector of a segment.
An altitude of a triangle is the perpendicular segment from a vertex to the line containing
the opposite side.
AM  BC
Definition of altitude of a triangle.
A median of a triangle is a segment from a vertex to the midpoint of the opposite side.
YD  ZD
Definition of median of a triangle.
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Complementary angles are two angles whose measures have the sum of 900.
•
•
RST and TSU are complementary angles.
m RST + m TSU = 90 
Definition of complementary angles.
Supplementary angles are two angles whose measures have the sum of 1800.
A pair of supplementary angles also can also be referred to as a linear pair.
A linear pair is when two angles together form a straight line.
•
•
•
RST and TSU are supplementary angles.
m RST + m TSU = 180 
RST and TSU are a linear pair.
Definition of supplementary angles.
If two angles are supplementary, then the angles form a linear pair.
Conversely,
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If two angles form a linear pair, then they are supplementary.
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There are other reasons for proofs that have been developed from the definition of
supplementary angles.
Supplements of congruent angles are congruent.
If A  B ,
then C  D .
If two angles are congruent and supplementary, then each is a right angle.
If A  B ,
then A and B are right angles.
Vertically opposite angles are congruent.
•
•
•
1 and 3 are vertically opposite angles.
2 and 4 are vertically opposite angles.
1  3
2  4
Definition of vertical angles.
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An isosceles triangle is a triangle with two congruent sides.
JK  JL
Definition of isosceles triangle.
Two properties of an isosceles triangle are used in proofs.
In the same triangle, or in congruent
triangles, angles opposite congruent sides are
congruent.
In the same triangle, or in congruent
triangles, sides opposite congruent angles are
congruent.
Parallel lines have many properties that are used in proving triangles congruent.
This diagram shows the various parts of parallel lines with AB
transversal.
CD and t being the
If two parallel lines are cut by a transversal, then the alternate
interior angles are congruent.
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•
•
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3  6
4  5
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If two parallel lines are cut by a transversal, then the
corresponding angles are congruent.
•
•
1  5
3  7
•
•
2  6
4  8
If two parallel lines are cut by a transversal, then the same side
interior angles are supplementary.
•
•
3  5  180
4  6  180
Proving Triangles Congruent
The congruence postulates outline what is needed in order to prove that two triangles are
congruent. The information that you are given will determine which congruence postulate
that you use.
Use the outline for problem solving that was shown previously. The process will seem
much simpler if you take a logical approach to each proof.
•
•
•
•
Read the problem.
Develop a plan.
Carry out the plan.
Look back.
Some of the proofs will show how this pattern has been used. In later proofs, it will be
assumed that the pattern has been followed.
Example
Given:
ABC with CM bisecting ACB
and AC  BC
Prove:
ACM  BCM
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Solution
Read the problem.
Given:
ABC with CM bisecting ACB
and AC  BC
Prove:
ACM  BCM
Develop a plan.
Mark the congruent parts on the two triangles.
A bisector bisects an angle.
Common side.
AC  BC
ACM  BCM
CM  CM
Determine the congruence postulate that can be applied.
(Side)
(Angle)
(Side)
SAS Postulate
Carry out the plan.
1.
2.
3.
4.
5.
Statement
Reason
AC  BC
CM bisects ACB
ACM  BCM
CM  CM
ACM  BCM
1.
2.
3.
4.
5.
Given
Given
Definition of angle bisector.
Reflexive Property
SAS Postulate
Look Back
The final statement of the proof will always be what was originally asked to be proven. In
this example, the last step was that the triangles were in fact congruent, and this was
determined by using the SAS Postulate.
Example
Given:
Figure with AB  BD , ED  BD
C is the midpoint of BD
Prove:
ABC  EDC
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Solution
Read the problem.
Given:
Figure with AB  BD , ED  BD
C is the midpoint of BD
Prove:
ABC  EDC
Develop a plan.
Mark the right angles on the two triangles.
D and B
(Angle)
A midpoint bisects a segment.
DC  BC
(Side)
Mark the vertical angles.
DCB and BCA
(Angle)
Determine the congruence postulate that can be applied.
ASA Postulate
Carry out the plan.
Statement
Reason
1.
2.
AB  BD , ED  BD
D and B are right angles.
1.
2.
3.
4.
5.
6.
7.
D  B
C is the midpoint of BD
DC  BC
DCB  BCA
ABC  EDC
3.
4.
5.
6.
7.
Given
Perpendicular lines meet to form right
angles.
Any two right angles are congruent.
Given
Definition of midpoint.
Definition of vertical angles.
ASA Postulate
Look Back
The final statement of the proof will always be what was originally asked to be proven. In
this example, the last step was that the triangles were in fact congruent, and this was
determined by using the ASA Postulate.
The way in which you label the corresponding parts is very important. It all depends
on how the two triangles correspond so that they are congruent.
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Most of the time it is important to label angles using three letters. Sometimes in a
diagram the angles are labelled with numbers. In this case refer to a certain angle in its
numbered form.
Example
Given:
ABC with RC  SC , 5  6
Prove:
BCR  ACS
Solution:
Read the problem.
Given:
ABC with RC  SC , 5  6
Prove:
BCR  ACS
If it is easier to see, separate the triangles.
Develop a plan.
Mark the congruent sides on the two triangles.
Common angle.
Mark the congruent angles.
RC  SC
C  C
5  6
(Side)
(Angle)
(Angle)
•
5 and 6 are not in the triangles that are to be proven, so it is necessary to see if
this information can lead to other corresponding parts being congruent.
•
It does follow that 7 and 8 are supplementary angles of the two congruent angles
and therefore congruent as well.
Determine the congruence postulate that can be applied.
ASA Postulate
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Carry out the plan.
Statement
Reason
1.
2.
3.
Given
Given
Definition of a linear pair.
4.
RC  SC
5  6
5 and 7 are supplementary.
6 and 8 are supplementary.
7  8
4.
5.
6.
C  C
BCR  ACS
5.
6.
Supplements of congruent angles are
congruent.
Reflexive property
ASA Postulate
1.
2.
3.
Example
Given:
Figure ABCD with AD  DC , CB  AB ,
DC BA
Prove:
ADC  CBA
Solution
Read the problem.
Given:
Figure ABCD with AD  DC , CB  AB ,
DC
BA
Prove:
ADC  CBA
Develop a plan.
Mark on the diagram everything that is given.
•
•
•
The common side is congruent.
Two right angles are formed by the perpendicular lines and are therefore congruent.
From the two segments that are parallel, the alternate interior angles are
congruent.
Determine the congruence postulate that can be applied.
AAS Postulate
The side is not included between the two congruent angles.
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Carry out the plan.
Statement
Reason
1.
2.
AD  DC , CB  AB
ADC and CBA are right angles
1.
2.
3.
ADC  CBA
3.
4.
5.
DC BA
ACD  CAB
4.
5.
6.
7.
CA  AC
ADC  CBA
6.
7.
Example
Given:
Figure ABCD with
AD  DC , CB  AB ,
CD  AB
Prove:
ADC  CBA
Given
Perpendicular lines meet to
form right angles.
Any two right angles are
congruent.
Given
If two lines are parallel, the
alternate interior angles are
congruent.
Reflexive property
AAS Postulate
Solution
Read the problem.
Given:
Figure ABCD with
AD  DC , CB  AB ,
CD  AB
Prove:
ADC  CBA
Separating the triangles,
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Develop a plan.
Mark on the diagram everything that is given.
•
•
•
Two right angles are formed by the perpendicular lines and are therefore congruent.
This makes these triangles right triangles.
The common side is congruent. This is also the hypotenuse that is congruent.
Two segments are congruent. These are the legs of the right triangle that are
congruent.
Determine the congruence postulate that can be applied.
HL Theorem
This theorem can only be applied when the triangles involved are right triangles.
Carry out the plan.
Statement
Reason
1.
2.
AD  DC , CB  AB
ADC and CBA are right angles
1.
2.
3.
4.
5.
6.
ADC and CBA are right triangles
CD  AB
CA  AC
ADC  CBA
3.
4.
5.
6.
Given
Perpendicular lines meet to
form right angles.
Definition of a right triangle
Given
Reflexive property
HL Theorem
When asked these types of questions, here are some points to remember.
•
•
•
•
Determine which triangles you want to prove congruent. Draw a diagram to
separate the triangles, if that gives you a better idea of which parts are
corresponding.
Use the diagram to mark the congruent parts, or right angles.
Take the information that is given to you and use each piece of information to
determine as much as you can about the congruence of corresponding parts.
The last statement in the two-column proof should be the same as what the
question asked you to prove.
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Proving Corresponding Parts of Congruent Triangles are
Congruent
In the previous section you proved that two triangles were congruent using the different
congruence postulates.
You have also learned that:
If the corresponding parts of two triangles are congruent,
then the triangles are congruent.
The converse is also true and will be the main focus of this section.
If two triangles are congruent,
then the corresponding parts of the triangles are congruent.
You will now be asked to prove that either sides or angles of two triangles are congruent.
The following steps will be followed:
•
•
First prove the two corresponding triangles are congruent.
State the corresponding parts (sides or angles) are congruent using the following
reason:
Corresponding parts of congruent triangles are congruent.
•
The abbreviated form CPCTC will be used in this course.
Example
Given:
Prove:
Mathematics C30
Figure with 1  2 and 3  4
AB  CB
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Preparation Package
First prove the two corresponding triangles are congruent.
Statement
Reason
1.
1  2; 3  4
1.
Given
2.
3.
BD  BD
ABD  CBD
2.
3.
Reflexive Property
ASA postulate
Secondly, the corresponding sides are now congruent.
4.
AB  CB
4.
CPCTC
Example
Given:
Isosceles ABC with AB  AC ,
Median AD
Prove:
BAD  CAD
Solution
Read the problem.
Given:
Isosceles ABC with AB  AC ,
Median AD
Prove:
BAD  CAD
Develop a plan.
Determine the two triangles that contain the two angles that are to be proven congruent.
•
BAD and CAD are the two triangles that first need to be proven congruent.
Mark on the diagram everything that is given.
•
•
•
Two congruent sides are AB  AC .
The common side is congruent.
The median of a triangle goes to the midpoint of BC . It follows that BD  CD .
Determine the congruence postulate that can be applied.
SSS Postulate
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Preparation Package
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Because the two triangles are proven congruent, the corresponding parts of these triangles
are then also congruent.
Carry out the plan.
1.
2.
3.
4.
5.
6.
Statement
Reason
AB  AC
AD is the median
BD  CD
AD  AD
BAD  CAD
BAD  CAD
1.
2.
3.
4.
5.
6.
Given
Given
Definition of median
Reflexive property
SSS Postulate
Corresponding parts of congruent
triangles are congruent.
The point to remember when proving that corresponding parts are
congruent is to first prove the triangles congruent that contain the
corresponding parts.
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Preparation Package