pom10smcoursereviewfinal

Course Review Note: Length and angle measures may vary slightly due to rounding. Course Review Question 1 Page 438 a) Let l represent the length and w represent the width, then 2l + 2w = 40.

b) If n represents one number and q represents the other number, then 152

n q+= .

c) If q represents the number of quarters and l represents the number of loonies, then

0.25q + l = 37. d) If a represents the number of adult tickets sold and s represents the number of student tickets

sold, then 20a + 12s = 9250. Course Review Question 2 Page 438 a) The point of intersection is (3, –1). b) The point of intersection is (–2, –5). c) The point of intersection is (2, 2).

MHR • Principles of Mathematics 10 Solutions 1

Course Review Question 3 Page 438 a) x + 4y = 6 x = –4y + 6 2x – 3y = 1 Substitute –4y + 6 for x into equation .

( )2 3 1

2 38 12 3 1

11

1

111

x yy

y yyy

− =

− += −

− =−

=

2 15 5

1

y x

4 6y − =+−

6 3

Substitute y = 1 into equation .

( )2 3 1

2 3 112 4

2

x yx

xx

− =

− =

==

The solution to the linear system is x = 2 and y = 1. b) y = 6 – 3x y = 2x + 1 Substitute 6 – 3x for y into equation . 2 1

xxx

= += +

− = −=

x−

Substitute x = 1 into equation .

( )2 1

12 13

y x= +

= +

= The solution to the linear system is x = 1 and y = 3.

2 MHR • Principles of Mathematics 10 Solutions

c) 5x – y = 4 3x + y = 4 y = 4 – 3x Substitute 4 – 3x for y into equation .

( )5 4

5 45 4 3

8 84

1

x y

x xxx

x− =

=

− + ==

−

=

y x= −

= −

=

4 3x−

Substitute x = 1 into equation .

1

( )4 34 31

The solution to the linear system is x = 1 and y = 1. Course Review Question 4 Page 438 a) x + y = 55 2x – y = –4 3x = 51 + x =17 Substitute x = 17 into equation .

( )17555538

x yyy

+ =

+ =

= Check by substituting x = 17 and y = 38 into both original equations. In x + y = 55: In 2x – y = –4:

( ) ( )35517 8x y= +

= +

=

L.S. 55=R.S.

( ) ( )2

7 382 14

x y= −

= −

= −

L.S. 4= −R.S.

L.S. = R.S. L.S. = R.S. The solution checks in both equations. The solution to the linear system is x = 17 and y = 38.


b) 2a + b = 5 a – 2b = 10 4a + 2b = 10 = × 2 a – 2b = 10 5a = 20 + a = 4 Substitute a = 4 into equation .

( )2 5

2 4

3− 3

53

a bbb

+ =

+ =

= −

2a b= +

= +

=

L.S.24

a b

Check by substituting a = 4 and b = –3 into both original equations. In 2a + b = 5: In a – 2b = 10:

( ) ( )2

54

5=R.S.

( ) ( )2

10

L.S. = −

= −

=

10=R.S.−

L.S. = R.S. L.S. = R.S. The solution checks in both equations. The solution to the linear system is a = 4 and b = –3. c) 4k + 3h = 12 4k – h = 4 4h = 8 – h = 2 Substitute h = 2 into equation .

( )24 3 12

4 3 124 6

1.5

k hk

kk

+ =

+ =

==

Check by substituting k = 1.5 and h = 2 into both original equations. In 4k + 3h = 12: In 4k – h = 4:

( ) ( )12=R.S.4 3

4 312

.5 21

k h= +

= +

=

L.S.

)( ) (4

.5 24 14

k h= −

= −

=

L.S. 4=R.S.

L.S. = R.S. L.S. = R.S. The solution checks in both equations. The solution to the linear system is k = 1.5 and h = 2.


d)5a – 2b = 5 3a + 2b = 19 8a = 24 + a = 3 Substitute a = 3 into equation .

( )5 2 5

5 2 52 1

50

a bbbb

− =

− =

− = −=

3

Check by substituting a = 3 and b = 5 into both original equations. In 5a – 2b = 5: In 3a + 2b = 19:

( ) ( )2

5523

55a b= −

= −

=

L.S. 5=R.S.

( ) ( )3 23 2 519

3a b= +

= +

=

L.S. 19=R.S.

L.S. = R.S. L.S. = R.S. The solution checks in both equations. The solution to the linear system is a = 3 and b = 5. Course Review Question 5 Page 438 The lines have the same slope, but a different y-intercept. The lines are parallel and they have no point in common. Course Review Question 6 Page 438 a) The point of intersection is (6.7, 1.7). b) The point of intersection is (–4.4, –2.3) c) The point of intersection is (–0.1, –0.9).


Course Review Question 7 Page 438 2a + 3b = 124 3a – 2b = 180 – 124 3a – 2b = 56 Multiply equation by 3 and by 2. Then, subtract. 6a + 9b = 372 × 3 6a – 4b = 112 × 2 13b = 260 b = 20 Substitute b = 20 into equation .

( )202 3 124

2 3 1242 64

32

a ba

aa

+ =

+ =

==

The value for a is 32 and the value for b is 20. Course Review Question 8 Page 438

The speed up the river was 60 ,5

or 12 km/h.

The speed down the river was 60 ,3

or 20 km/h.

Let b represent the speed of the boat. Let r represent the speed of the river. b + r = 20 b – r = 12 2b = 32 + b = 16 Substitute b = 16 into equation . b + r = 20 16 + r = 20 r = 20 – 16 r = 4 The speed of the boat is 16 km/h, and the speed of the river is 4 km/h.


Course Review Question 9 Page 438 Let s represent the required volume of 60% hydrochloric acid. Let t represent the required volume of 30% hydrochloric acid.

125s t+ = 0.6 0.3 0.36 1250.6 0.3 45

s ts t+ = ×+ =

Multiply equation by 0.6. Then, subtract equation . 0.6s + 0.6t = 75 × 0.6 0.6s + 0.3t = 45 0.3t = 30 t = 100 Substitute t = 100 into equation .

10012512525

s ts

s

+ =+ =

=

The required volume of 60% acid is 25 mL, and the required volume of 30% acid is 100 mL.


Course Review Question 10 Page 438

2 1 23 5

2 115 15 15 23 55 10 3 3 30

5 3 37

x y

x y

x yx y

− ++ =

− +× + × = ×

− + + =+ =

( )

2 5 27 3

2 521 21 21 27 33 6 7 35 42

3 7 13

x y

x y

x yx y

+ +− = −

+ +× − × = −

+ − − = −− = −

Multiply equation by 3, and equation by 5. Then, subtract. 15x + 9y = 111 × 3 15x – 35 y = –65 × 5 44y = 176 y = 4 Substitute y = 4 into equation .

( )45 3 3

5 3 35 2

5

x yx

xx

+ =

+ =

==

775

The solution to the linear system is x = 5 and y = 4.


Course Review Question 11 Page 439 Line segment AB:

( )

( )

1 2 1 2

2 6 1

, ,2 2

,2 2

2,3

5

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

=

−

( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

6 –2 1

B

5

A

8 4

80

x x y y= − + −

⎡ ⎤= − + −⎣ ⎦

= +

=

Line segment CD:

( )

( ) ( )

( )

1 2 1 2, ,2 2

,2 2

5,

7 –− 3 4 –

0

4

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞+ +

= ⎜ ⎟⎝ ⎠

= −

( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

2

2 2

CD

4 8

80

x x y y= − + −

⎡ ⎤= + −⎣ ⎦

= + −

=

2

3 7 4 4− −− −

Line segment EF:

( )

( )

( )

1 2 1 2, ,2 2

,2 2

2, 1

1 –22 6

.5

x x y yx y

−−

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎝ ⎠

= −

( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

EF

– –

8 1

65

6 –2

2 –1

x x y y= − + −

⎡ ⎤ ⎡ ⎤= +⎣ ⎦ ⎣ ⎦

= + −

=

−


Course Review Question 12 Page 439 a) Let the midpoint of KL be M.

( )

( )

( )

1 2 1 2

1 43

, ,2 2

,2 2

0, 2 5

3

.

x x y yx y

− −−

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎝ ⎠

= −

( )

2 1JM

2 1

.6 5

y ymx x−

=−

=

=

4 2.01

5−−−

The y-intercept is 2.5. The equation of the line through JM is 6.5 2.5y x= − . b) Let the midpoint of JL be N.

( )

( )

( )

1 2 1 2

4 –

, ,2 2

,2 2

2,0

41 3

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎝ ⎠

=

( )( )

2 1KN

2 1

– –– –

0.

13

02

2

y ymx x−

=−

=

=

y x bb

bb

= +

− = − +− =

( )0.20

1 0.60.4

.21 3= +−−

The equation of the median from vertex K is 0.2 0.4y x= − .


c) From part b), the midpoint of JL is N(2, 0).

2 1JL

2 1

4 43 14

y ymx x−

=−

=

= −

−−

−

The slope of the right bisector of JL is 14

.

( )14

14

0 = +2

102

12

y x b

b

b

b

= +

= +

− =

The equation of the right bisector of JL is 1 14 2

y x= − .


a) ( ) ( )

( ) ( )

( ) ( )

22 1 2 1

2 2

2 2

41 10 18 63

AC

31 45

54.6

2x x y y= − + −

= − + −

= + −

( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

B

41 87 18 30

C

46 12

47.5

x x y y= − + −

= − + −

= − + −

Fire station B is closer. b) Answers will vary. For Example: Plot points A, B, and C. Use the Measure menu to calculate

the distances.



( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

2 10

DE

8 4

80

4 0

x x y y= − + −

= − + −

= − +

=

( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

8 2 6

EF

10 10

4

200

x x y y= − + −

= − + −

= − + −

=

− −

( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

2 2

FG

14 2

200

6

x x y y= − + −

⎡= −⎣

= + −

=

)2 2

8 8 6⎤ ⎡ ⎤+ −⎦ ⎣ ⎦− − −

( ) (

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

6 1

D

0 8 0

G

4 8

80

x x y y= − + −

= − + −

= − + −

=

−

Adjacent sides are equal in length. DEFG is a kite. Course Review Question 15 Page 439 a)


b) The coordinates of M are The coordinates of N are

( )

( )

( )

1 2 1 2

10 62

, ,2 2

,62 2

4,2

x x y yx y + +⎛= ⎜⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎝ ⎠

=

⎞⎟ ( )

( )

1 2 1 2, ,2 2

,2 2

8,8

2 14 10 6

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

=

−

c) ( ) ( )

( ) ( )

( ) ( )

22 1 2 1

2 2

2 2

8 4

MN

4 6

5

2

2

2 13

8

2x x y y= − + −

= − + −

= +

=

=

( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

22

2 2

14 6 6

KL

8 12

6

208

4 13

x x y y= − + −

⎡ ⎤= − + −⎣ ⎦

= +

=

=

−

The length of MN is half the length of KL.

d) 2 1MN

2 1

6432

y ym

8 28 4

x x−

=−

=

=

=

−−

( )

2 1KL

2 1

12832

y ym

6 614 6

x−

−

x

=

−

=−

=

=

−

The slopes are the same. MN is parallel to KL.


Course Review Question 16 Page 439 Let M be the midpoint of QR. The coordinates of M are

( )

( )

1 2 1 2

2 4 5

, ,2 2

,2 2

1,3

1

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

=

( )

2 1QR

2 1

− 1 54 2

4623

y ymx x−

=−

=−

= −

= −

−−

The coordinates of M are (1, 3). The slope of the right bisector of QR is 32

.

( )32

32

3 = +

32

1

y x b

b

b

= +

=

( )

3 32 23 32 2

3−

32

y x= +

= +

= −≠ −

The coordinates of point P do not satisfy the equation of the right bisector of the line segment QR. The point P(–3, –2) does not lie on the right bisector of line segment QR.



a) 2AB

2 1

1

88

1

y ym

2 644

x x−

=−

=

−−

=

=

− −−−

( )( )

2 1BC

2 1

36

21

y ymx x−

5 22 4− −−

=−

=− −

−=

= −

( )

2 1CD

2 1

991

y ym

4 511 2

x x−

=−

=

=

=

−−−

2 1AD

2 1

4 611 4

27

y ym −=

x x−−

=−

= −

One pair of opposite slopes are equal, but the others are not. ABCD is a trapezoid. b) Answers will vary. For example: Plot the given points and construct line segments joining

them. Use the Measure menu to measure the slopes of the line segments.


Course Review Question 18 Page 439 a) Let I be the point at which the connector should

be located.

The shortest distance is the perpendicular distance HI to WM, so point I should lie on the perpendicular to line segment WM that passes through point H(24, 22).

2 1WM

2 1

2 342 01

328

4

y ymx x−

=−

=−

−=

−=

2y x b

bb

y x

= +

− == −

−

2 = +

( )46

4 6

4

The slope of HI is 14

− .

( )

14

22 24

−

1

84

2

y x b

b

b

= +

= − +

=

The equation of the line passing through points M(2, 2) and W(10, 34) is 1 284

y x= − + .

Substitute 4x – 6 for y in the equation 1 284

y x= − + .

14 6 24

16 24 11217 136

8

x x

x xxx

− = − +

− = − +==

86

26

y x

( )4 64

= − 8= −

=

The water main should be located at point I(8, 26).


b) ( ) ( )

( ) ( )

( ) ( )

2 2HI 2 1 2 1

2 2

2 2

24

16 4

27216.5

8 22 26

d x x y y= − + −

= − + −

= + −

=

Since each grid interval represents 0.5 m, the connection requires about 0.5 × 16.5, or 8.25 m

of pipe. Course Review Question 19 Page 439 a) The radius of the circle is 7. 2 2 49x y+ = b) ( )25 + =

)

( )2

2 2

61

6

6

1x y+ =

c) ( ( )228 + =

8

2 2

67

6

3

7x y+ =

1

A r= π

= π

0


2 2 64x y+ = The radius of the circle is 8 units. The diameter is 16 units.

( )

2

2

20

The area of the circle is about 201 square units. Course Review Question 21 Page 440

2 2 2

2

602 360

42

s sss

+ =

=

The side length of the chute is about 42 cm.


Course Review Question 22 Page 440 a) The centroid is the point where the three medians of a triangle intersect. b) Determine the equation of two of the medians of the triangle and then find the point of

intersection of these two lines. c) Answers will vary. For example: Construct the triangle. Construct the midpoints of the sides.

Join the midpoints to the vertices to form the medians. Measure the coordinates of the point of intersection of the medians.

Course Review Question 23 Page 440 a) Answers will vary. Draw any triangle. Draw a median. Find the length of the base and height

of each of the two triangles formed. Calculate the area of each triangle. b) Answers will vary. For example: Construct a triangle. Construct a median. Construct triangle

interiors in the two triangles formed. Measure the area of each triangle using the Measure menu.


( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

6 –

AC

12 4

16

6 14 10

0

x x y y= − + −

⎡ ⎤= − + −⎣ ⎦

= +

=

( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

6 2 14 2

BC

4 12

160

x x y y= − + −

= − + −

= +

=

The length of AB equals the length of BC. ΔABC is isosceles.


Course Review Question 25 Page 440 a) Answers will vary. For example:

2 1DE

2 1

4 148 210653

y ymx x−

=−

=

−=

= −

−−

2 1EF

2 1

10 418 86

1035

y ymx x−

=−−

=−

=

=

The slopes of DE and EF are negative reciprocals. ΔDEF is a right triangle. b) Answers will vary. For example: Another way to show that ΔDEF is a right triangle is to show

that the sides satisfy the Pythagorean theorem. Course Review Question 26 Page 440 a)


b) From the diagram, the midpoint of JL is X(2, 2), the midpoint of LK is Y(5, 3), and the midpoint of KJ is Z(–1, –5).

( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

5

XY

3 1

10

2 3 2

x x y y= − + −

= − + −

= +

=

( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2

JK

6 2

40

2 10

x x y y= − + − 2 2

4 4 6⎡ ⎤ ⎡ ⎤+ −= −⎣ ⎦ ⎣ ⎦−− −

= +

=

=

)

( ) (

( ) ( )

( ) ( )

2 22 1 2 1

2 2

YZ

6 8

10010

2 21 5 5 3

x x y y= − + −

=

= − + −

==

− + −− −

( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

2 2

8 4

JL

12 16

4002

0 6

0

1

x x y y= − + − 2 2

⎡ ⎤ ⎡ ⎤− −= − + −⎣ ⎦ ⎣ ⎦

= +

==

)

( ) (

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

XZ

3 7

58

1 2 5 2

x x y y= − + −

= − + −

− + −

−−

=

=

) ( ) (

( ) ( )

( ) ( )

2 22 1 2 1

22

2 2

8 2 10

KL

6 14

4

232

2 58

x x y y= − + −

⎡ ⎤= − + −⎣ ⎦

= +

=

=

−

Corresponding sides are in the same proportion of 2:1. ΔXYZ is similar to ΔJKL.


Course Review Question 27 Page 440 a) From the diagram, the midpoint

of JL is X(3, 1), the midpoint of LK is Y(7, 4), and the midpoint of KJ is Z(5, 5).

2 1JL

2 1

0152

2

2

14

y ymx x−

=−−

=−

= −

= −

53

y x bb

b

= +

− =

The slope of the right bisector of JL is 2.

( )22

1= +

The equation of the right bisector of JL is 2 5y x= − .

2 1LK

2 1

842

y ym

8 09 5

x x−

=−

=

=

=

−−

The slope of the right bisector of LK is 12

− .

( )

12

4 7

−

1

12

25

y x b

b

b

= +

= − +

=

The equation of the right bisector of LK is 1 12 2

y x 5= − + .


2 1JK

2 1

6834

y ym

8 29 1

x x−

=−

=

=

=

−−

The slope of the right bisector of JK is 43

− .

( )

43

5 5

−

4

33

35

y x b

b

b

= +

= − +

=

The equation of the right bisector of JK is 4 33 3

y x= − +5

5

.

b) Use the method of substitution to find the circumcentre.

2y x= −1 12 2

y x 5= − +

Substitute 2x – 5 for y in equation .

1 12 52 2

4 10 155 25

5

x x

x xxx

− = − +

− = − +==

5

( )2 5

52 55

y x= −

= −

=

Two of the right bisectors intersect at (5, 5). Check to see that (5, 5) also lies on the third bisector.

( )4 35 4 33 3 3

5

53

x− + = − +

=

5

The coordinate (5, 5) satisfies the equation of the third right bisector. All three intersect at the circumcentre C(5, 5).


c) ( ) ( )

( ) ( )

( ) ( )

22 1 2 1

2 2

2 2

JC

4 3

255

5 1 5 2

2x x y y= − + −

= − + −

= +

==

( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

CK

4 3

255

9 5 8 5

x x y y= − + −

= − + −

= +

==

( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

5 5

CL

25

0

0 5

5

5

x x y y= − + −

= − + −

= + −

==

The circumcentre of ΔJKL is equidistant from its vertices. Course Review Question 28 Page 441 a) Squares and rectangles have diagonals that are equal in length. b) Squares, rhombi, and parallelograms have diagonals that bisect each other. c) Squares, rhombi, and kites have diagonals that meet at right angles. Course Review Question 29 Page 441 a) Answers will vary. For example: Draw a parallelogram. Find the midpoints of two opposite

sides. Join the midpoints. Find the length of the line segment created, and the lengths of the other two sides of the parallelogram.

b) Answers will vary. For example: Construct a parallelogram. Construct the midpoints of two

opposite sides. Join the midpoints. Measure the length of the line segment created, and the lengths of the other two sides of the parallelogram.



a) 2 1AB

2 1

2 72 254

54

y ymx x−

=−

=

−=−

=

−−−

( )

2 1BC

2 1

4623

y ym

2 24 2

x x−

=−

− −=

−=

= −

− −

( )

2 1CD

2 1

54

y ym

3 28 4

x x−

−=

−

=

=

−−

2 1AD

2 1

3287

3

4

26

y ymx x−

=−

=

−=

= −

)

−−

Opposite sides are parallel. Adjacent sides are not perpendicular. ABCD is a parallelogram.

b) ( ) (

( ) ( )

( ) ( )

22 1 2 1

2 2

2 2

4

A

2

C

2

2

9

5

7

8

2

x x y y= − + −

= − −

= + −

=

+

)

−

( ) (

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

BD

10 1

101

8 2 3 2

x x y y= − + −

⎡ ⎤= − + −⎣

=

=

− ⎦

+

The diagonals are not equal in length.

c) Find the midpoint of each side. For AC: For BD:

( )

( )

1 2 1 2

7 22 4

( , ) ,2 2

,2 2

3,2.5

x x y yx y + +⎛= ⎜⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎝ ⎠

=

⎞⎟

( )

1 2 1 2( , ) ,2 2

,2 2

3,2

2 8 2 3

.5

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

=

−−

The midpoints of the diagonals are the same. The diagonals bisect each other.

d) 2AC

2 1

1

92

y ym

2 74 2

x x−

=−

=

= −

−−−

( )

2 1BD

2 1

11

23 2

8

0

y ymx x−

=−

=

−−

=−

The slopes are not negative reciprocals. The diagonals are not perpendicular.



3 03

x yy x

− + == +

2y x= −

1 52

y x= − +

2 12 0

1 62

x y

y x

+ + =

= − −

There are two pairs of equal slopes, of 1 and 12

− . They are not negative reciprocals. No sides are

perpendicular. The quadrilateral is a parallelogram. Course Review Question 32 Page 441

a) ( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

4 2 2 5

PC

2 3

13

x x y y= − + −

= − + −

= + −

=

( ) ( )

( ) ( )

( ) ( )

2 22 1 1

22

2 2

QC

2 3

2 1

3

4

1

6

x 2x y y

−

= − + −

⎡ ⎤= − + −⎣ ⎦

= − +

=

)

( ) (

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

4

R

7

C

3 2

13

2 4

x x y y= − + −

= − + −

= − + −

=

All three points are equidistant from C. They lie on a circle with centre at C. b) Calculate the midpoint for PQ.

( )

( )

1 2 1 2( , ) ,2 2

,2

5 12 62

4,2

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎝ ⎠

=

−

This is the centre of the circle. The centre of the circle lies on the right bisector of chord PQ.


Course Review Question 33 Page 441 Answers will vary. For example: Plot the three given points. Join the points with line segments to form two chords. Find the midpoints of the chords. Construct perpendicular lines through the midpoints. The perpendicular lines meet at the centre of the circle. Course Review Question 34 Page 441 a) The relation is neither linear nor

quadratic. x y First Differences

Second Differences

–2 –12 –1 –3 9

0 0 3 –6 1 –3 –3 –6 2 –12 –9 –6 3 –36 –24 –15

b) The first differences are constant.

x y First Differences

–2 –6 –1 –3 3 0 0 3 1 3 3 2 6 3 3 9 3

The relation is linear. c) The first differences are constant.

x y First Differences

–2 –12 –1 –7 5

0 –2 5 1 3 5 2 8 5 3 13 5

The relation is linear. Course Review Question 35 Page 441 a) The ball was released from a height of 2.1 m. b) Use graph paper or a graphing calculator to graph the relation. The

maximum height reached by the ball was about 4.43 m. c) 2

( ) ( )2 8.25 2.11 1+ +

7.3 8.25 2.1

7.33.05

h t t= − + +

= −

= The height of the hoop was 3.05 m.


Course Review Question 36 Page 441 a) b)

c) d)


Course Review Question 37 Page 442 a) b) The height is 0 m at t = 0 s and t = 11 s. The maximum height occurs at t = 5.5 s. ( )

( ) ( )5.5 5.5

5 11

5 1

151

h t t= − −

⎡ ⎤= − −⎣ ⎦1

2

3

35

24 3

2

2 3

y ax

aa

a

y x

= +

− = +− =

= − +

The maximum height is approximately 151 m. c) The maximum height occurs after 5.5 s. d) Answers will vary. For example: The lava will be ejected away from the crater and so it will

probably fall on land that is below the crater. The length of time in the air will probably be more than 11 s.

Course Review Question 38 Page 442 a) The vertex is (0, 3). b)

( )

2 ( )( )25 = +− (7 2a )( )

( )( )2

3 5

3 57 71

3 5

2 15

2

x x

aa

y x x

x x

y a= − +

= − +−

− = −=

= − +

= + − Course Review Question 39 Page 442 The horizontal distance to the vertex from the given x-intercept is 9. The other x-intercept is the same distance past the vertex, at –11.



a) b) 08 =1 1 133

− = c) ( ) 5 1232

−− = −

d) 22 2

5 4

−⎛ ⎞ =⎜ ⎟⎝ ⎠

5 e) ( )018 1− = f) 14 3

3 4

−⎛ ⎞ =⎜ ⎟⎝ ⎠

Course Review Question 41 Page 442 a) At 500ºC, there are 10 doublings. 102

1024t ==

It would take the wood 1024 s to burn. b) At 650ºC, there are 5 halvings.

51

21

32

t ⎛ ⎞= ⎜ ⎟⎝ ⎠

=

It would take the wood 132

s to burn.

Course Review Question 42 Page 442 a) ( ) ( )3 4 5 6 3 12 5 3

8 18x x x x 0

x− + + = − + +

= +

b) ( ) ( )6 3 2 5 6 18 2 14 28

a a a aa

0+ − − = + − +

= +

c) ( ) ( )4 1 3 2 4 4 12 2 4

2 8k k k

k− − − = − − +

= −

k d) ( ) ( ) 2 2

2

2 3 4 2 5 6 8 2 5

8 3

t t t t t t t t

t t

− + + = − + +

= −

e) ( ) ( )1 36 3 8 10 3 1.5 6 7.52 4

3 9

p p p

p

+ − − = + − +

= − +

p

f) ( ) ( )2 2 3 2 3

3

3 1 2 3 4 3 3 3 2 3

7

y y y y y y y y y y y y

y y

− − − − + = − − − + −

= −

2 4


Course Review Question 43 Page 443 a)

( )( ) ( )( ) ( )( )( )( )

2 2 2

2

2

2 3 2 2 2 1 3 2 1

2 3 2 6 2 4 2 2 6 3

2 5 5 5

10 10 10

SA x x x x x x

x x x x x x x x x

x x

x x

⎡ ⎤= + − + − + + + +⎣ ⎦

= + − − + − + − + + + +

= + −

= + −

10

16

b)

( ) ( )

2

2

10 10 10

1 5 50 10290

SA x x= + −

= + −

= The surface area of the box is 290 cm2. Course Review Question 44 Page 443 a) b) ( )2 24 8x x x+ = + + ( )( ) 24 4 1y y y 6− + = − c) ( ) d) 2 25 10a a a− = − + 25 ( )( ) 23 1t 3 1 9 1t t+ − = − e) ( )( ) 25 3 5 3 25 9a b a b a b+ − = − 2 f) ( ) ( )2 2

2

2 3 1 2 9 6 1

18 12 2

m m m

m m

+ = + +

= + +


Course Review Question 45 Page 443 a) ( )( ) ( )2 2 2

2

3 3 4 9 8

2 8 7

m m m m m m

m m

− + + − = − + − +

= − +

16

2

b) ( ) ( )( )2 2 2

2

3 2 1 2 3 1 3 1 12 12 3 18 2

30 12 1

t t t t t t

t t

+ + − + = + + + −

= + + c) ( )( ) ( )2 2 2 2

2 2

2 3 2 3 2 3 3 18 8 27 18 3

9 18 11

x y x y x y x y x xy y

x xy y

+ − − − = − − + −

= − + −

d) ( ) ( ) ( )( )2 2 2 2

2

2 1 2 1 2 1 2 4 4 1 4 4 1 4

9 4

y y y y y y y y

y

− + + − − + = − + + + + − +

= +

2y

2

2

e) ( ) ( )2 2 2 2 2

2 2

4 2 2 16 8 2 8 8

18 9

m n m n m mn n m mn n

m n

+ + − = + + + − +

= + f) ( ) ( )( )2 2 2 2

2 2

5 2 5 3 4 3 4 3 20 100 125 48 27

68 100 98

t z t z t z t tz z t z

t tz z

− + − + = − + + −

= − + Course Review Question 46 Page 443 a) ( )5 35 5 7k k− = − b) ( )24 20 4 5h h h h− = − c) ( )22 8 2 1 4xy xy xy y− = − d) ( )( )2 25 5 5x x x− = + − e) ( )( )21 49 1 7 1 7m m− = − + m f) ( )

( )(

2 2 2 24 16 4 4

4 2 2

a b a b

a b a b

− = −

= − + )


Course Review Question 47 Page 443 a) The length is n – 2 and the width is n – 3. b) ( ) ( )

( ) ( )

( )

2

2

2 2 3

2 2 3

22

5 65 6

3

8 8

80

8

P n n

A n n

⎡ ⎤= − + −⎣ ⎦⎡ ⎤= − + −⎣ ⎦

=

= − +

= − +

=

The perimeter is 22 cm and the area is 30 cm2. Course Review Question 48 Page 443 a) ( )( )2 12 4 3x x x x− − = − + b) ( )( )2 3 18 6 3y y y y+ − = + − c) ( )( )2 11 24 8 3m m m m+ + = + + d) ( )( )2 8 15 5 3t t t t− + = − − e) f) 2 3 4 cannot be factored.x x+ + ( )( )2 13n 40 8 5n n n− + = − − g) ( )( )2 30 6 5w w w w− − = − + h) ( )( )214 5 7 2m m m m+ − = − + Course Review Question 49 Page 443 a) b) ( 22 10 25 5x x x+ + = + ) ( )22 12 36 6y y y− + = − c) d) 2 6 16 cannot be factored.m m+ + ( )224 12 9 2 3x x x+ + = + e) f) ( )22 225 20 4 5 2r rs s r s− + = − ( )

( )

2 2 2

2

5 20 20 5 4 4

5 2

2x xy y x xy y

x y

− + = − +

= −

Course Review Question 50 Page 443 Answers will vary.


Course Review Question 51 Page 443 a) b) ( )22 8 16 4x x x+ + = + ( )224 12 9 2 3x x x− + = −

The only possible value for p is 9. ( )22 8 16 4x x x− + = − Possible values for p are 8 and –8. c) ( )2225 40 16 5 4x x x+ + = + The only possible value for p is 25. Course Review Question 52 Page 443

( )( )

2 2 2121

m nm n m n

− =

− + =

The only pairs of integers whose product is 21 are (1, 21), (3, 7), (–1, –21), and (–3, –7). Possible values for (m, n) are thus (11, 10), (5, 2), (–11, –10), and (–5, –2). Course Review Question 53 Page 444 a)

( )( )

2

2

2

4 1

4 4 4

2 3

y x x

x x

x

= + +

= + + − +

= + −

1

5

3

The vertex is (–2, –3), and the axis of symmetry is x = –2. b)

( )( )

2

2

2

6 5

6 9 9

3 4

y x x

x x

x

= − − −

= − + + − −

= − + +

The vertex is (–3, 4), and the axis of symmetry is x = –3. c)

( )( )

2

2

2

3 4

4 4 4

2 7

y x x

x x

x

= − −

= − + + − +

= − + +

The vertex is (–2, 7), and the axis of symmetry is x = –2.


Course Review Question 54 Page 444 a) The minimum is (–1.5, 0.5). b) The maximum is (0.3, 1.3). c) The minimum is (–0.3, –3.7). Course Review Question 55 Page 444 a)

( )(

2 2 33 1

y x xx x

= + −

= + − )

)

)

The x-intercepts are –3 and 1. b)

( )(

2 6 55 1

y x xx x

= + +

= + + The x-intercepts are –5 and –1. c)

( )( )

2 4 42 2

y x xx x

= − +

= − − The x-intercept is 2. d)

( )(

24 12 92 3 2 3

y x xx x

= − +

= − − 2 3 0

1.5x

x− =

= The x-intercept is 1.5.



( )( )

2 3 28 07 4 0

7 or 4

x xx x

x x

+ − =

+ − =

= − =

( ) ( )

2

2

3 28

3 87 20

7

x x= + −

= + −

0=R.S.

( ) ( )

2

2

3 28

=

L.S.

−− 4 3 20

4 8

x x= + −

= + −

=

L.S.

=

L.S.

0=R.S.

L.S. = R. S. L.S. = R. S. The roots are –7 and 4. b)

( )( )

2 7 10 05 2 0

5 or 2

m mm m

m m

+ + =

+ + =

= − = −

( ) ( )

2

2

7 10

7 05 10

5

m m= + +

= + +

0=R.S.

( ) ( )

2

2

7 10

7 0−− 2 2 10

m m= + +

= + +

=

L.S.

62

n=

=

=

L.S. n= −

= −

−−

9− 9

0=R.S.

L.S. = R. S. L.S. = R. S. The roots are –5 and –2. c)

( ) ( )( ) ( )( )( )

( ) ( )

2

2

2

2

2 27 152 15 27 0

2 18 3 27 0

2 18 3 27 0

2 9 3 9 0

9 2 3 0

9 0 or 2 3 09 or 1.5

n nn n

n n n

n n n

n n n

n n

n nn n

= −

+ − =

+ − − =

+ − + =

+ − + =

+ − =

+ = − =

= − =

( )

2

2

2

21

( )27 1527 15162

−

=

R.S.

4.5

n=

=

=

L.S.7 15

4.5

n

( )1.5 .5

2

2

2

2 ( )27 152 1

= −

= −

=

R.S.

L.S. = R. S. L.S. = R. S. The roots are –9 and 1.5.


d)

( ) ( )

( ) ( )( ) ( )( )( )

( ) ( )

2

2

2

2

3 4 4 1 0

3 12 4 4 03 7 4 0

3 3 4 4 0

3 3 4 4 0

3 1 4 1 0

1 3 4 0

1 0 or 3 4 041 or 3

k k k k

k k k kk k

k k k

k k k

k k k

k k

k k

k k

− + + + =

− + + + =

− + =

− − + =

− − − =

− − − =

− − =

− = − =

= =

( ) ( )

( ) ( ) ( ) ( )1 1

3 4 4 1

3 4 1 14 1

0

k k k k= − + + +

⎡ ⎤ ⎡= − + +⎣ ⎦ ⎣=

L.S.

⎤+ ⎦

0=R.S.

L.S. = R. S. ( ) ( )3 4 4 1

3 4 4 1⎤+ ⎥⎦

8 4 74 43

4 4 4 43 3 3 3

3 332 4 283 3 3

0

k k k k= − + + +

⎡ ⎤ ⎡⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞= − + +⎜ ⎟⎝ ⎠

=

L.S. 0 =R.S.

L.S. = R. S.

The roots are 1 and 43

.


Course Review Question 57 Page 444 a) Answers will vary. For example: ( )( )

2

5 2

3 10

y x x

x x

= − +

= − −

b) Answers will vary. For example: ( )( )

2

4 3 3 2

12 17 6

y x x

x x

= + +

= + +


( ) ( )( )

2

25 − =−

k

20

2

4 0

4 025 8 0

258

b a

k

c

k

k

− =

− =

=

k

− =

− =

− == ±

4 0

4 0400 1 0

2500

4

k

b ac

kk

− =

− =

− ==

b)

( ) ( )( )

2

2

2

1 9

4 0

4 0

36 06

b ac

k

c)

( ) ( )( )

2

2


( )

( )( )

2

5 36

5 36 09 4 0

x x

x xx x

− =

− − =

− + =

The negative root is inadmissible. The dimensions are 9 cm by 9 – 5, or 4 cm.



a)

( ) ( ) ( )( )( )

2 42

2

1 172

b b acxa

− ± −=

−

±=

=2 41 1 1 4

1± −−− −

b)

( ) ( ) ( ( ))( )

2 42

2

2 6014

1 157

b b aca−

−

±=

=±

=

k − ±=

22 4 27

2 7± − −− −

c) d)

( ) )( )(( )

2 42

48 82

8 884

4 222

b b acxa

− ± −=

− ± −=

− ±=

− ±=

2 2 32

− ( ) )( )(( )

2

4 4

42

42

4 968

1 62

b b aca−

− ±

− ±=

− ±=

h − ±=

2 4 54− −

=

e)

( ) )( )(( )

2 42

44 42

4 286

2 73

b b acaa

− ± −=

− ±=

− ±=

−±

=

2 3 13− −

−


Course Review Question 61 Page 444 a) Let R represent the revenue. Let x represent the number of $10 decreases in price. ( )( )

( )( )( )

2

2

2

200 10 90 5

18 000 100 500 400 100 50

0 50 8 2

0 50 4 2

R x x

17 600 x xx x

x x

x x

= − +

= + −

= + −

= + −

= − +

The negative root gives a price increase. The value for x is 4. The number of jackets sold is 90 + 5(4), or 110. The selling price for a revenue of $17 600 is 200 – 10(4), or $160. b) ( )( )

( )( )( )

2

2

2

200 10 90 5

18 000 100 500 2400 1

R x x

15 60000 50

0 50 48 2

0 50 8 6

x xx x

x x

x x

= − +

= + −

= + −

= + −

= − +

The negative root yields a higher selling price. The value for x is 8. The number of jackets sold is 90 + 5(8), or 130. The lowest selling price that gives a revenue of $15 600 is 200 – 10(8), or $120. Course Review Question 62 Page 444 Let x represent the length of one leg of the triangle.

( )

( )( )( )

22 2

2 2

2

2

28 20

784 56 4002 56 384 0

2 28 192 0

2 12 16 0

x x

x x xx x

x x

x x

+ − =

+ − + =

− + =

− + =

− − =

The sides of the triangle measure 12 cm and 16 cm.


Course Review Question 63 Page 444 Let x represent the width of the deck. ( )( )

2

2

10 2 4 2 135

40 28 4 13595 28 4 0

x x

x xx x

+ + =

+ + =

− + + =

( ) ( )( )( )

2

228 28 4

42

42

95

28 23048

28

4

488

b b acxa

− ± −=

− ± −=

− ±=

− ±=

−

The negative root is inadmissible. The width of the deck is 2.5 m. Course Review Question 64 Page 445 ΔADE ~ ΔACB ∠A is common ∠ADC =∠ACB (corresponding angles). Course Review Question 65 Page 445

4.31 0.2

0.2 4.321.5

h

hh

=

==

The height of the tree is 21.5 m.



a) 8.5tan A13.2

A 33

=

∠ °

b) 1.2tan A0.7

A 60

=

∠ °

Course Review Question 67 Page 445 tan 0.08

4.6θθ=

∠ °

The angle of inclination of the hill is approximately 4.6 . ° Course Review Question 68 Page 445

a) cos5312.612.6cos537.6

x

xx

° =

= °

The length of side x is about 7.6 m.

b) sin 2812.512.5sin 285.9

x

xx

° =

= °

The length of side x is about 5.9 cm.



a) sin335656sin3330

b

bb

° =

= °

cos335656cos3347

a

aa

° =

= °

A 90 3357

∠ = ° − °= °

In ΔABC, = 57°, a = 47 cm, and b = 30 cm. A∠

b) tan 416060 tan 4152

e

ee

° =

= °

60cos41

60cos4180

d

d

d

° =

=°

F 90 4149

∠ = ° − °= °

In ΔDEF, F 49 , 79 m, and 52 m.d e∠ = ° = =

c) 10sinT15

T 42

=

∠ °

U 90 42

48∠ = ° − °

= °

2 215 1011

uu

2= +

In ΔUST, ∠ = T 42 , U 48 , and u = 11 m.° ∠ = °

d) 8tan P13

P 32

=

∠ °

R 90 32

58∠ = ° − °

= °

2 213 815

qq

2= +

In ΔQRP, P 32 , R 58 , and 1 m. 5 c q∠ = ° ∠ = ° =



2 2

3.5sin Z4.8

Z 47

Y 90 4743

4.8 3.5 XZXZ 3.3

=

∠ °

∠ = ° − °= °

= + 2

2

rom

In ΔXYZ, Z 47 , Y 43 , and XZ 3.3 cm.∠ = ° ∠ = ° = Course Review Question 71 Page 446 sin A 0.5

A 30

B 90 3060

=∠ = °

∠ = ° − °= °

Course Review Question 72 Page 446 a) 2 27.5 14.8

16.6ll= +

The coast guard boat is approximately 16.6 km fthe yacht.

b) 7.5tan B14.8

B 26.9

=

∠ °

The coast guard boat must travel at an angle of approximately 26.9° south of west to reach the yacht.


Course Review Question 73 6

tan 26.6100

100 tan 26.6 tan 26.6100tan26.6tan26.6

hx

x hhx

° =+

° + ° =− °

=°

tan71.7

tan71.7

tan71.7

hx

x hhx

° =

° =

=°

100 tan 26.6tan 26.6 tan71.7

tan71.7 100 tan71.7 tan 26.6 tan 26.6tan71.7 tan 26.6 100 tan71.7 tan 26.6

100 tan71.7 tan 26.6tan71.7 tan 26.6

60.0

h h

h hh h

h

h

− °=

° °° − ° ° = °

° − ° = ° °° °

=° − °

The height of the bridge is approximately 60.0 m. Course Review Question 74 Page 446

2 254 2861

2865 61

61 182030

vv

x

xx

= +

=

=

2

The length of side x is about 30 cm.



sin A sin B

sin A sin

13.6sin A 15.3sin4815.3sin 48sin A

13.6sin A 0.8360..

4815.3 13.6

°

.A 57

a b=

=

= °°

=

=∠ °


sinC sin B

sin sin54

65. 3 .286=

° °sin38.2 54sin65.6

54sin65.6sin38.2

79.5

c b

c

c

c

c

=

° = °°

=°

The length of side c is about 79.5 cm.



a) sin R sin P

sin R sin

43sin R 38sin5138sin51

5138 43

°

sin R43

sin R 0.6867...R 43.4

r p=

=

= °°

=

=∠ °

Q 180 43.4 5185.6

∠ = ° − ° − °= °

sinQ sin P

sin sinsin51 43sin85.

4385.6 51°

643sin85.6

sin5155.2

q p

q

q

q

q

=

=

°°

=°

= °°

In ΔQRP, R 43.4 , Q 85.6 , and 55.2 cm.q∠ = ° ∠ = ° = b) J 180 75 32

73∠ = ° − ° − °

= °

sin K sin J

sin sinsin73 20.5sin7

20.575 73° °

520.5sin75

sin7320.7

k j

k

k

k

k

=

=

°°

=°

= °

sin L sin J

sin sinsin73 20.5sin3

20.532 73°

220.5sin32

sin7311.4

l j

l

l

l

l

=

=

°°

=°

= °°

In ΔJKL, J 73 , 20.7 cm, and 11.4 cm.k l∠ = ° = =



ABC 32 4375

C 180 75105

A 105

sin sinsin105 8.5sin 43

8.5sin 43sin105

6.0

sin sin

8.543 105

8.532 105

=

°°

° °sin105 8.5sin32

8.5sin32sin105

4.7

x

x

x

x

y

y

y

y

∠ = ° + °= °

∠ = ° − °= °

∠ = °

=

° = °°

=°

° = °°

=°

The sides of the parallelogram measure approximately 4.7 cm and 6.0 cm. Course Review Question 79 Page 446

( )( )( )( )°

2 2 2

2 2 220 10

2

20 10 40

cosS

2 cos13.9

s r t rt

ss

= + −

= + −

The length of side s is about 13.9 m.



( )( )

2 2 2

2 2 23.2 2.8

cos L2

cos L2

cos L 0.6601...L 48.7

l m kmk

− −=

−− −

=−

=∠ °

2.53.2 2.8

( )( )

2 2 2

2 2 23.2 2.5

cos K2

cos K2

cos K 0.5406...K 57.3

k m lml

− −=

−− −

=−

=∠ °

2.83.2 2.5

M 180 57.3 48.7

74.0∠ = ° − ° − °

= °

In ΔKLM, L 48.7 , K 57.3 , and M 74.0 .∠ = ° ∠ = ° ∠ = ° Course Review Question 81 Page 447

( )( )( )

( )( )( )24.6

12

5 2

.1 1

4.6

7.4 19

5 12

.82

24.6

.1 24.65

5

17.4 24.65 19

1

8

0

.

4

s

A s s a s b s c

+ +=

=

= − − −

= − − −

The area of the triangle is approximately 104 m2.



1071 38° °

180 38B2

71

sin B sin A

sin sinsin38 10sin71

10sin71sin3

15.48

b a

b

b

b

b

° − °∠ =

= °

=

=

° = °°

=°

The perimeter of the triangle is about 15.4 + 15.4 + 10, or 40.8 cm. Course Review Question 83 Page 447

a) tan 2750

50 tan 27 tan 2750tan 27tan 27

hx

x hhx

° =+

° + ° =− °

=°

tan35

tan35

tan35

hx

x hhx

° =

° =

=°

50 tan 27tan 27 tan35

tan35 50 tan35 tan 27 tan 27tan35 tan 27 50 tan35 tan 27

50 tan35 tan 27tan35 tan 27

93.6

h h

h hh h

h

h

− °=

° °° − ° ° = °

° − ° = ° °° °

=° − °

The length of side h is about 93.6 m.


b) tan 47200

200 tan 47 tan 47200tan 47tan 47

hx

x hhx

° =−

° − ° =− + °

=°

tan 42

tan 42

tan 42

hx

x hhx

° =

° =

=°

200 tan 47tan 47 tan 42

tan 42 200 tan 42 tan 47 tan 47tan 42 tan 47 200 tan 42 tan 47

200 tan 42 tan 47tan 42 tan 47

97.9

h h

h hh h

h

h

− + °=

° °− ° + ° ° = °

− ° − ° = − °° °

=° + °

°

The length of side h is about 97.9 m. Course Review Question 84 Page 447 After 45 min, the first boat travelled 0.75 × 10, or 7.5 km, and the second boat travelled 0.75 × 8, or 6.0 km. The angle between the bearings is 79 47 , or 32 .° − ° °

( )( )( )2 2 27.5 6.0 7.2 cos4.0

dd= + − 5 6.0 32°

The distance between the boats after 45 min was about 4.0 km.


Course Review Question 85 Page 447 a) F 180 58 53

69∠ = ° − ° − °

= °

sin F sin D

sin sin8

69 58° °sin58 8sin69

8sin69sin5888.

f d

f

f

f

f

=

=

° = °°

=°

sin E sin D

sin sin8

53 58° °sin58 8sin53

8sin53sin5857.

e d

e

e

e

e

=

=

° = °°

=°

In ΔDEF, F 69 , 8.8 cm, and 7.5 cm.f e∠ = ° = = b)

S 180 46 7361

sinS sin R

sin sinsin73 8sin61

8sin61sin

861 73° °

737.3

s r

s

s

s

s

∠ = ° − ° − °= °

=

=

° °°

=°

=

sinT sin R

sinT sin

8sinT 6sin736sin73sinT

8sinT 0.7172...

T 46

736 8

°t r

=

=

= °°

=

=∠ °

In ΔRST, T 46 , S 61 , and 7.3 m.s∠ = ° ∠ = ° =


c) ( )( )( )( )2 cos5 7 68°

2 2 2

2 2 25

2 cosA

6.7

9

a b c ab

aa

=

−

+ −

= +

( )( )

2 2 2

256.9 7

− −2 26

c

.9 7

os B2

cos B2

cos B 0.7413...B 42

b a cac

− −=

−

=−

=∠ °

C 180 42 6870

∠ = °− ° −= °

°

In ΔABC, B 42 , C 70 , and 6.9 cm.a∠ = ° ∠ = ° =

d)

( )( )

2 2 2

2 2 210

co

14

sW2

cosW2

cosW 0.625W 51

w x yxy

− −=

−

− −=

−

=∠ °

( )(11

10 14 )

2 2 2

210 11

cos Y2

cos Y2

cos Y 0.1136...Y 83

y x wxw

− −=

−− −

=−

=

2 21410 11

∠ ° X 180 51 83

46∠ = ° − ° − °

= ° In ΔWXY, W 51 , Y 83 , and X 46 .∠ = ° ∠ = ° ∠ = ° Course Review Question 86 Page 447 The largest angle is opposite the longest side, 18 cm. Let be the largest angle. A∠

( )( )2 2 217 15cosA2

cosA 0.3725...A 68

− −=

−

=∠ °

1817 15

The largest angle is 68°.


pom10smcoursereviewfinal

Documents