Chapter 8 Trigonometry of Acute Angles - Weeblypetrinmath.weebly.com/uploads/6/2/5/6/6256684/pom10_solutions_ch8...Chapter 8 Trigonometry of Acute Angles ... sinP sin 7.5sinP 5.2sin68

Chapter 8 Trigonometry of Acute Angles Note: Slightly different answers may be obtained if measures are calculated in a different order. Chapter 8 Get Ready Chapter 8 Get Ready Question 1 Page 394

a)

sin X =4.08.1

cos X =7.08.1

tan X =4.07.0

b) 4.0tan X7.0

X 30

Z 90 30Z 60

=

∠ °

∠ = ° − °∠ = °

Chapter 8 Get Ready Question 2 Page 394 a)

k 2 = 3.92 + 5.22

k = 6.5 cm

sin M =3.96.5

cos M =5.26.5

tan M =3.95.2

b)

tan M =3.95.2

∠M 37°

B 90 3753

∠ = ° −= °

°

Chapter 8 Get Ready Question 3 Page 395 Answers may vary. For example: You could use the Pythagorean theorem or apply the trigonometric ratio for sin T.

MHR • Principles of Mathematics 10 Solutions 1

Chapter 8 Get Ready Question 4 Page 395

2 2 2

2 2

1.6tan 44

tan 44 1.61.6

tan 441.7

2.3

R 90 4

1.6 1.7

446

rr

r

r

l m rll

° =

° =

=°

= +

= +

∠ = ° − °= °

2

In ΔLMR, r is 1.7 cm, l is 2.3 cm, and∠R is 46°, to the specified rounding. Chapter 8 Get Ready Question 5 Page 395 a)

b)

2 2 2

2 2

7.2

5.3tan F4.8

F 48

W 90 4

5.3

842

4.8p w fpp

= +

= +

=

∠ °

∠ = ° − °= °

2

In ΔPWF, p is 7.2 km, ∠ F is 48°, and ∠W is 42°, to the specified rounding.

2 MHR • Principles of Mathematics 10 Solutions


a)

tan35° =BY25

BY = 25tan35°BY 17.5

The measure of BY is approximately 17.5 m.

b)

tan 20° =AY25

AY = 25tan 20°AY 9.1

The height of the taller building is approximately 17.5 + 9.1, or 26.6 m.


tan 25° =40d

d tan 25° = 40

d =40

tan 25°d 86

The ship is approximately 86 m from the base of the cliff. Chapter 8 Get Ready Question 8 Page 395

a) b)

y = mx + by − mx = mx + b− mx

y − mx = b

s =dt

s × t =dt× t

st = d

c)

P = 4sP4=

4s4

P4= s

d) P = a + b+ cP − a − c = a + b+ c − a − cP − a − c = b


Chapter 8 Get Ready Question 9 Page 395 a)

a sin B( )= b sin A( )a sin B( )

sin B=

b sin A( )sin B

a =b sin A( )

sin B

b) c2 = a2 + b2

c2 − a2 = a2 + b2 − a2

c2 − a2 = b2

c)

A = 6s2

A6=

6s2

6A6= s2

±A6= s

d) x

sin X=

ysin Y

xsin X

sin Y =y

sin Ysin Y

x sin Y( )sin X

= y


Chapter 8 Section 1 The Sine Law Chapter 8 Section 1 Question 1 Page 402 a)

∠A = 180° − 71° − 53°= 56°

asin A

=b

sin Ba

sin

=sin

471°56°

a sin71° = 4sin56°

a =4sin56°sin71°

a 4 In ΔABC, a is approximately 4 cm.

b)

esin E

=f

sin Fe

sin

=sin

1370°51°

esin70° = 13sin51°

e =13sin51°

sin70°e 11

In ΔDEF, e is approximately 11 m.


Chapter 8 Section 1 Question 2 Page 402

a)

asin A

=b

sin Ba

sin

=sin

2.544°62°

a sin 44° = 2.5sin62°

a =2.5sin62°

sin 44°a 3.2

In ΔABC, a is approximately 3.2 cm. b)

E 180 59 7249

sin E sin D

sin sinsin59 5.7sin

5.749 95 °°

495.7sin 49

sin59.05

e d

e

e

e

e

∠ = ° − ° − °= °

=

=

° = °°

°=

In ΔDEF, e is approximately 5.0 mm.



a)

sin Yy

=sin X

xsin Y

=sin65°

1713

sin Y =13sin65°

17sin Y 0.6931∠Y 44°

b)

sinCc

=sin K

ksinC

=sin73°

2.21.9

sinC =1.9sin73°

2.2sinC 0.8259∠C 56°


a)

sin Hh

=sinG

gsin H

=sin 47°

454sin H = 5sin 47°

sin H =5sin 47°

4sin H 0.9142∠H 66°

b)

sin Tt

=sinS

ssin T

=sin72°

1.81.51.8sin T = 1.5sin72°

sin T =1.5sin72°

1.8sin T 0.7925∠T 52°


Chapter 8 Section 1 Question 5 Page 402 a)

L 180 45 7065

sin M sin P

sin sin18

70 45

1865 4sin 5

=

=

° °

° °

18sin70sin 45

24

sin L sin P

sin18sin65

sin 4523

m p

m

m

m

l p

l

l

l

∠ = ° − ° − °= °

=

°=

°

=

°=

°

In ΔLPM, ∠L is 65°, m is 24 m, and l is 23 m, to the specified rounding. b)

∠D = 180° − 57° − 72°= 51°

dsin D

=f

sin Fd

sin

=sin

2072°51°

d =20sin51°

sin72°d 16

esin E

=f

sin Fe

sin=

sin2072°57°

e =20sin57°

sin72°e 18

In ΔEFD, D is 51°, d is 16 cm, and e is 18 cm, to the specified rounding. ∠



a)

sin Dd

=sin H

hsin D

=sin73°

3429

sin D =29sin73°

34sin D = 0.8157∠D 55°

W 180 55 7352

∠ = ° − ° − °= °

sin W sin H

sin sin34sin52

sin

3452 73° °

7328

w h

w

w

w

=

=

°=

°

In ΔHDW, D is 55°, ∠W is 52°, and w is 28 cm, to the specified rounding. ∠

b)

sinQq

=sin R

rsinQ

=sin61°

1211

sinQ =11sin61°

12sinQ = 0.8017∠Q 53°

P 180 53 6166

∠ = °− ° − °= °

psin P

=r

sin Rp

sin

=sin

1261°66°

p =12sin66°

sin61°p 13

In ΔPQR, Q is 53°, P is 66°, and p is 13 m, to the specified rounding. ∠ ∠



a)

sin Rr

=sin K

ksin R

=sin68°

1513

sin R =13sin68°

15sin R 0.8036∠R 53°

A 180 53 6859

∠ = °− ° − °= °

sin P sin R

sin sin13sin59

sin

1359 53° °

5314

a r

a

a

a

=

=

°=

°

In ΔAKR, R is 53°, A is 59º, and a is 14 mm, to the specified rounding. ∠ ∠ b)

∠J = 180° − 57° − 48°= 75°

fsin F

=j

sin Jf

sin

=sin

2375°48°

f =23sin 48°

sin75°f 18

usin U

=j

sin Ju

sin=

sin2375°57°

u =23sin57°sin75°

u 20 In ΔUJF, J is 75°, f is 18 km, and u is 20 km, to the specified rounding. ∠


Chapter 8 Section 1 Question 8 Page 402 Answers will vary. Draw a triangle in the workspace. Measure lengths and angles using the Measure menu. Drag the vertices of the triangle to match the given information. Chapter 8 Section 1 Question 9 Page 402 a)

∠B = 180° − 46° − 74°= 60°

asin A

=b

sin Ba

sin

=sin

2.260°46°

a sin60° = 2.2sin 46°

a =2.2sin 46°

sin60°a 1.8

The length of the pole is approximately 1.8 m.

b)

csinC

=b

sin Bc

sin

=sin

2.260°74°

csin60° = 2.2sin74°

c =2.2sin74°

sin60°c 2.4

The length of the cable is approximately 2.4 m.



sin P sinT

sin P sin

7.5sin P 5.2sin685.2sin68sin P

7.5sin P 0.6428

P 40

J 180 40 6872

sin J sinT

sin sin

685.2 7.5

7.572 68

=

°

° °sin68 7.5sin72

7.5sin72sin68

7.7

p t

j t

j

j

j

j

=

=

= °°

=

=∠ °

∠ = ° − ° − °= °

=

° = °°

=°

The plane is approximately 7.7 km from the tower.



a)

sin Bb

=sinC

csin B

=sin50°

141814sin B = 18sin50°

sin B =18sin50°

14sin B 0.9849∠B 80°

The second wire makes an angle of approximately 80º with the ground.

b) sin801414sin8014

h

hh

° =

= °

The height of the pole is approximately 14 m.

c)

cos80° = x14

x = 14cos80°x 2

cos50° = y18

y = 18cos50°y 12

The base of the shorter wire is approximately 2 m from the pole. The base of the longer wire is approximately 12 m from the pole.

d) Answers will vary. For example: You can solve this problem using primary trigonometric

ratios.



A 180 60 5466

sin B sin A

sin sinsin66 150sin54

150sin54sin66

133

sin60133133

15054 66°°

sin60115

b a

b

b

b

b

h

hh

∠ = ° − ° − °= °

=

=

° = °°

=°

° =

°=

The valley is approximately 115 m deep. Chapter 8 Section 1 Question 13 Page 403

∠A = 180° − 70° − 55°= 55°

bsin B

=a

sin Ab

sin=

sin5.0

55°55°bsin55° = 5.0sin55°

b =5.0sin55°

sin55°b = 5.0

sin70° =h

5.0h = 5.0sin70°

4.7 The altitude of the aircraft is approximately 4.7 km.


Chapter 8 Section 1 Question 14 Page 403 Answers will vary. For example, since ABC is an isosceles triangle, either b = c or b = a. Case 1: b = 15 cm Case 2: b = 11 cm

sinCc

=sin B

bsinC

sin A

a=

sin Bb

sin A

=sin68.5°

15=

sin68.5°1115

∠C = 68.5°

∠A = 180° − 2 68.5°( )= 73°

11∠A = 68.5°

∠C = 180° − 2 68.5°( )∠C 43°=

There is no triangle with the given dimensions. In Case 1 the shortest side of the triangle will be opposite the greatest angle, and in Case 2 the longest side of the triangle will be opposite the least angle. Both of these situations are impossible. Chapter 8 Section 1 Question 15 Page 403

sin80.150.15sin80.021

cos80.150.15cos80.149

tan120.1490.149 tan120.032

x

xx

d

dd

y

yy

° =

= °

° =

= °

° =

= °

The height of the tower is approximately 21 + 32, or 53 m.


Chapter 8 Section 1 Question 16 Page 404 a) From Example 3 on page 400, a = 1600 km, b 1478 km, and c 1855 km.

s =12

a + b + c( )

=12

1600 +1478 +1855( )= 2466.5

A = s s − a( ) s − b( ) s − c( )=

2466.5( )2466.5−1600 2466.5−1478( ) 2466.5−1855( )1 136 610

The area of the Bermuda triangle is approximately 1 136 610 km2. b) Answers will vary. Chapter 8 Section 1 Question 17 Page 404

∠B = 180° − 55° − 58°= 67°

asin A

=b

sin Ba

sin

csinC

=b

sin Bc

sin

55°=

sin6.2

67°=

sin6.2

67°a sin67° = 6.2sin55°

a =6.2sin55°

sin67°a 5.5

58°csin67° = 6.2sin58°

c =6.2sin58°

sin67°c 5.7

6.2 5.5 5.7

28.7

s + +=

=

( )( )( )

( )( )( )8.7 8.7 6.2 8.7 5

14.4

.5 8.7 5.7

A s s a s b s c= − − −

= − − −

The area of the garden is approximately 14.4 m2. Chapter 8 Section 1 Question 18 Page 404 Answers will vary.


Chapter 8 Section 1 Question 19 Page 404 The Sine Law does not work if you replace the sines with cosines or tangents. Chapter 8 Section 1 Question 20 Page 404 a) Let the right angle be angle B.

sin Aa

=sin B

b=

sinCc

sin Aa

=sin90°

b=

sinCc

sin Aa

=1b=

sinCc

sin A =ab

sinC =cb

b) Answers will vary. For example: It is faster to use the sine ratio when solving right triangles. Chapter 8 Section 1 Question 21 Page 404

s = a + b+ c2

A =a + b+ c

2a + b+ c

2− a

⎛⎝⎜

⎞⎠⎟

a + b+ c2

− b⎛⎝⎜

⎞⎠⎟

a + b+ c2

− c⎛⎝⎜

⎞⎠⎟

=a + b+ c

2a + b+ c − 2a

2⎛⎝⎜

⎞⎠⎟

a + b+ c − 2b2

⎛⎝⎜

⎞⎠⎟

a + b+ c − 2c2

⎛⎝⎜

⎞⎠⎟

=a + b+ c

2b+ c − a

2⎛⎝⎜

⎞⎠⎟

a + c − b2

⎛⎝⎜

⎞⎠⎟

a + b− c2

⎛⎝⎜

⎞⎠⎟

=14

a + b+ c( ) a + b− c( ) b+ c − a( ) c + a − b( )


Chapter 8 Section 1 Question 22 Page 404 The given information is represented below.

B +C = 17A+ B = 12C + A = 15

C – A = 5 (Subtracting equation 1 from equation 2.) 2C = 20 (Adding equation 4 and equation 3.) C = 10 10 + A = 15 (From equation 3.) A = 5 Vahpav’s score is 2(5), or 10. The answer is B. Chapter 8 Section 1 Question 23 Page 404

a + b+ c2

= 105

b+ a + c2

= 106

c + a + b2

= 125

a + b+ c + 2a + 2b+ 2c2

= 336

2 a + b+ c( )= 336

a + b+ c3

= 56

The answer is C.


Chapter 8 Section 2 The Cosine Law Chapter 8 Section 2 Question 1 Page 409 a)

b2 = a2 + c2 − 2ac cos B( )b2 = 102 122

172 202

3.22 4.42 66°

+ − 2 10( )12( ) cos72°( )b 13

+ − 2 17( ) 20( ) cos39°( )c 13

+ − 2 3.2( ) 4.4( ) cos(

In ΔABC, b is approximately 13 cm. b)

c2 = a2 + b2 − 2ab cosC( )c2 =

In ΔABC, c is approximately 13 mm. Chapter 8 Section 2 Question 2 Page 409 a)

d 2 = e2 + f 2 − 2ef cos D( )d 2 = )

d 4.3 In ΔDEF, d is approximately 4.3 m. b)

z2 = x2 + y2 − 2xy cos Z( )z2 = 1.82 2.22 48°+ − 2 1.8( ) 2.2( ) cos( )

z 1.7

+ − 2 6.2( ) 4.8( ) cos(

In ΔYZX, z is approximately 1.7 cm. c)

p2 = m2 + n2 − 2mn cosC( )p2 = 6.22 4.82 54°)

p 5.1 In ΔMNP, p is approximately 5.1 mm.



u2 = v2 + t2 − 2vt cos D( )u2 = 1.42 1.82 52°+ − 2 1.4( )1.8( ) cos( )

u 1.4

+ − 2 1.1( )1.6( ) cos(

In ΔTUV, u is approximately 1.4 cm. b)

d 2 = e2 + f 2 − 2ef cos D( )d 2 = 1.12 1.62 74°)

d 1.7 In ΔDEF, d is approximately 1.7 km. Chapter 8 Section 2 Question 4 Page 409 a)

( )( )( )( )

2 2 2

2 2 213 15 13 15 70

7015 16

°

°

2 cos R

2 cos16

sinQ sin R

sinQ sin

16sinQ 15sin7015sin70sinQ

16sinQ 0.8810

Q 62

P 180 70 6248

r p q pq

rr

q r

= + −

= + −

=

=

= °°

=

∠ °

∠ = ° − ° −°

°=

In ΔPQR, P is 48°, Q is 62°, and r is 16 cm, to the specified rounding. ∠ ∠


b) ( )( )( )( )

2 2 2

2 2 226 24 26 24 47

4726 20

°

°

2 cos R

2 cos20

sin P sin R

sin P sin

20sin P 26sin 4726sin 47sin P

20sin P 0.9508

P 72

K 180 72 4761

r p k pk

rr

p r

= + −

= + −

=

=

= °°

=

∠ °

∠ = ° − ° −°

°=

In ΔKPR, P is 72°, K is 61°, and r is 20 m, to the specified rounding. ∠ ∠ c)

( )( )( )( )

2 2 2

2 2 2

2 cosA

2 cos10

sinC sin A

sinC sin

10sinC 9sin719sin71sinC

10si

8 9

nC 0.8510C 58

B 180 58 75

11

a b c bc

aa

c a

= + −

= + −

=

=

= °°

=

∠ °

∠ = ° − ° − °°=

8 9 71

719 10

°

°

In ΔABC, B is 51°, C is 58°, and a is 10 m, to the specified rounding. ∠ ∠



( )( )( )( )

2 2 2

2 2 2

2 cosG

2 s6

sin F sinG

sin F sin

6sin F 6sin636sin63sin F

6F 63

E 180 6

5 6

563 3

4

g e f ef

gg

f g

= + −

= + −

=

=

= °°

=

∠ °

∠ = ° − ° − °= °

co5 6 63

6366

°

°

In ΔEFG, F is 63°, E is 54°, and g is 6 cm, to the specified rounding. ∠ ∠ b)

( )( )( )( )

2 2 2

2 2 210 11 10 11 80

8011 14

°

°

2 cosX

2 cos14

sin Y sin X

sin Y sin

14sin Y 11sin8011sin80sin Y

14sin Y 0.7738

Y 51

W 180 51 8049

x w y wy

xx

y x

= + −

= + −

=

=

= °°

=

∠ °

∠ = ° − ° −°

°=

In ΔWXY, Y is 51º, W is 49º, and x is 14 m, to the specified rounding. ∠ ∠


Chapter 8 Section 2 Question 6 Page 410 Answers will vary. Use geometry software to draw a triangle. Use the Measure menu to measure angles and sides. Drag the vertices to model each of the triangles. Chapter 8 Section 2 Question 7 Page 410

a2 = b2 + c2 − 2ab cosA( )a2 = 752 902 38°+ − 2 75( ) 90( ) cos( )

a 56

+ − 2 2.0( ) 2.5( ) cos70°( )b 2.6

The length of the bridge is approximately 56 m. Chapter 8 Section 2 Question 8 Page 410 a)

b2 = h2 + s2 − 2hs cos B( )b2 = 2.02 2.52

Chandra’s home and school are approximately

2.6 km apart.

b)

sinS sin B

sinS sin

2.6sinS 2.5sin702.5sin70sinS

2.6sinS 0.9036

S 65

H 180 65 704

702.5 2.6

°

5

s b=

=

= °°

=

∠ °

∠ = ° − ° − °= °

The angle of elevation from home is 45° and the angle of elevation from school is 65°, to the specified rounding.


Chapter 8 Section 2 Question 9 Page 410 a2 = b2 + c2 − 2ab cosA( )a2 =

16.02 14.52

102 102

202 202 45°

+ − 2 16.0( )14.5( ) cos35°( )a 9.3

+ − 2 10( )10( ) cos 45°( )a 7.7

+ − 2 20( ) 20( ) cos(

The total length of the belt is approximately 14.5 + 16.0 + 9.3, or 39.8 cm. Chapter 8 Section 2 Question 10 Page 410 a) After 1 h, Wavedancer has travelled 10 nautical miles, and Ocean Princess has travelled 10 nautical miles.

a2 = b2 + c2 − 2ab cosA( )a2 =

After 1 h, the ships are approximately 7.7 nautical miles apart.

b) After 2 h, Wavedancer has travelled 20 nm, and Ocean Princess has travelled 20 nm.

a2 = b2 + c2 − 2ab cos A( )a2 = )

a 15.3

+ − 2 20( )10( ) cos 45°( )a 14.7

+ − 2 40( ) 20( ) cos(

After 2 h, the ships are approximately 15.3 nm apart.

c) If Wavedancer travels twice as fast, it will travel 20 nm in 1 h, and 40 nm in 2 h.

a2 = b2 + c2 − 2ab cos A( )a2 =

202 102

402 202 45°

a2 = b2 + c2 − 2ab cos A( )a2 = )

a 29.5

If Wavedancer travels twice as fast, after 1 h the ships will be approximately 14.7 nautical miles apart, and after 2 h they will be approximately 29.5 nautical miles apart.


Chapter 8 Section 2 Question 11 Page 410 a) ( )

( )( )( )61.5°

2 2 2

2 2 211.1 5.4 11.

2 cosA

1 5.42 cos9.8

a b c ab

aa

= + −

= −+

Sprockets B and C should be approximately 9.8 cm apart. b)

sin B sin A

sin B sin

9.8sin B 11.1sin61.511.1sin61.5sin B

9.8B 84.5

C 180 61.5 84.534.

61.511.1 9.8

°

15252 9152

0

b a=

=

= °°

=

∠ °

∠ = ° − ° − °= °

+ − 2 1525( ) 915( ) cos37°(

In ΔABC, B is 84.5° and ∠C is 34.0°, to the specified rounding. ∠ Chapter 8 Section 2 Question 12 Page 411 Answers will vary. Chapter 8 Section 2 Question 13 Page 411 a) b)

a2 = b2 + c2 − 2ab cosA( )a2 = )a 966

The thresholds are approximately 966 m apart.


Chapter 8 Section 2 Question 14 Page 411 You will fly a distance of 400 × 0.1, or 40 km.

a2 = b2 + c2 − 2ab cos A( )a2 = 452 402

102 102

+ − 2 45( ) 40( ) cos15°( )a 12

+ − 2 10( )10( ) cos104.5°( )a 15.8

You will be approximately 12 km from the storm cloud. Chapter 8 Section 2 Question 15 Page 411 Let C be the right angle.

c2 = a2 + b2 − 2ab cosC( )= a2 + b2 − 2ab cos90°( )= a2 + b2 − 2ab 0( )= a2 + b2

This is the Pythagorean theorem. Chapter 8 Section 2 Question 16 Page 411 Solutions for Achievement Checks are shown in the Teacher Resource. Chapter 8 Section 2 Question 17 Page 411 Answers will vary. Chapter 8 Section 2 Question 18 Page 411 a)

a2 = b2 + c2 − 2ab cos A( )a2 =

The hydrogen atoms are approximately 15.8 cm apart in the model. b) Since this is an isosceles triangle, the two unknown base angles are equal.

B C 180 104.52 B 75.5

B 37.75

∠ +∠ = ° − °∠ = °∠ = °

Angles B and C both measure 37.75°.


Chapter 8 Section 2 Question 19 Page 411 Answers may vary. For example: If the distances are measured between the centres of each pulley, the calculation of the length of the belt must also include the length of the belt which wraps around each pulley. Assume that the total distance of tapproximately one and a half times the complete pulley circumference.

he belt on each pulley is

C = πd= π 2.8( )

8.8

The length of the belt is approximately 39.8 + 1.5(8.8), or 53 cm. Chapter 8 Section 2 Question 20 Page 411 The number of ways that 12 players can be arranged on a bench is 12 ×11×10 × 9 × 8× 7 × 6 × 5× 4 × 3× 2 ×1 Put Sam and Nick next to each other, and treat them as one player. The number of arrangements is 2 ×11×10 × 9 × 8× 7 × 6 × 5× 4 × 3× 2 ×1 The initial factor of 2 takes into account that Sam and Nick can reverse positions. The probability that Sam and Nick are not beside each other is

12×11×10× 9× 8× 7 × 6× 5× 4× 3× 2×1− 2×11×10× 9× 8× 7 × 6× 5× 4× 3× 2×112×11×10× 9× 8× 7 × 6× 5× 4× 3× 2×1

=12− 2

12

=56

The correct answer is D.



sin A( )2 + cosA( )2 = oppositehypotenuse

⎛⎝⎜

⎞⎠⎟

2

+ adjacenthypotenuse

⎛⎝⎜

⎞⎠⎟

2

= opposite2 +adjacent2

hypotenuse2

= hypotenuse2

hypotenuse2

= 1


Chapter 8 Section 3 Find Angles Using the Cosine Law Chapter 8 Section 3 Question 1 Page 418

a)

cos A =a2 − b2 − c2

−2bc

cos A =

82 92 102− −−2 9( )10( )

cos A 0.6500∠A 49°

b)

cos B =b2 − a2 − c2

−2ac

cos B =

122 132 102− −−2 13( )10( )

cos B 0.4808∠B 61°

c)

cosC =c2 − a2 − b2

−2ab

cosC =

142 162 172− −−2 16( )17( )

cosC 0.6415∠C 50°



a)

cos K =k 2 − m2 − n2

−2mn

cos K =

132 142 162− −−2 14( )16( )

cos K 0.6317∠K 51°

b)

cos U =u2 − t2 − v2

−2tv

cos U =

2.42 1.8 2.5− 2 − 2

−2 1.8( ) 2.5( )cos U 0.4144∠U 66°

c)

cosG =g 2 − f 2 − h2

−2 fh

cosG =

4.82 5.12 6.22− −−2 5.1( ) 6.2( )

cosG 0.6548∠G 49°


a)

cos D =d 2 − a2 − r 2

−2ar

cos D =

2102 1702 1902− −−2 170( )190( )

cos D 0.3235∠D 71°

b)

cos W =w2 − h2 − n2

−2hn

cos W =

1.72 1.42 1.22− −−2 1.4( )1.2( )

cos W 0.1518∠W 81°



a)

cos J = j2 − m2 − v2

−2mv

cos J =

10.02 8.32 12.02− −−2 8.3( )12.0( )

cos J 0.5667∠J 55.5°

cos V =v2 − m2 − j2

−2mj

cos V =12.0 8.32 10.022 − −

−2 8.3( )10.0( )

cos M =m2 − j2 − v2

−2 jv

cos M =

8.3 10.0 12.022 − 2 −−2 10.0( )12.0( )

cos V 0.1499∠V 81.4°

cos M 0.7296∠M 43.1°

b) Methods will vary.

( )( )

2 2 2

2 2 28.3 12.0

cosJ2

cosJ2

cosJ 0.5667J 55.5

j m vmv

− −=

−− −

=−

∠ °

( )( )

2 2 2

10.08.3 12.0

2 212.08.3 10.0

28.3 10.0

cosV2

cosV2

cosV 0.1499V 81.4

v m jmj

− −=

−

− −=

−

∠ °

∠M = 180° − 55.5° − 81.4°= 43.1°

c) The answers are the same. Explanations may vary. For example: Calculations in the second

method are easier to make.



a)

cos U =u2 − t2 − v2

−2tv

cos U =

5 6 72 − 2 − 2

−2 6( ) 7( )cos U 0.7143∠U 44.4°

cos V =v2 − t2 − u2

−2tu

cos V =

72 62 52− −−2 6( ) 5( )

cos V = 0.2∠V 78.5°

T 180 44.4 78.5

57.1∠ = ° − ° − °

= °

b)

cos P =p2 − m2 − y2

−2my

cos P =

4.92 5.42 4.42− −−2 5.4( ) 4.4( )

cos P 0.5158∠P 59.0°

cos Y =y2 − m2 − p2

−2mp

cos Y =

4.42 5.42 4.92− −−2 5.4( ) 4.9( )

cos Y 0.6389∠Y 50.3°

M 180 59.0 50.3

70.7∠ = ° − ° − °

= °



a)

cos N =n2 − b2 − g 2

−2bg

cos N =

152 142 122− −−2 14( )12( )

cos N 0.3423∠N 70.0°

cosG =g 2 − b2 − n2

−2bn

cosG =

122 142 152− −−2 14( )15( )

cosG 0.6595∠G 48.7°

B 180 70.0 48.7

61.3∠ = ° − ° − °

= °

b)

cos D =d 2 − r 2 − t2

−2rt

cos D =

5.02 3.82 4.62− −−2 3.8( ) 4.6( )

cos D 0.3032∠D 72.3°

cos T =t2 − r 2 − d 2

−2rd

cos T =

4.62 3.82 5.0− − 2

−2 3.8( ) 5.0( )cos T 0.4811∠T 61.2°

R 180 72.3 61.2

46.5∠ = ° − ° − °

= °


Chapter 8 Section 3 Question 7 Page 418 Answers will vary. Use geometry software to draw a triangle. Use the Measure menu to measure side lengths and angles. Drag the vertices to model each triangle. Chapter 8 Section 3 Question 8 Page 419

a)

cosA =a2 − b2 − c2

−2bc

cosA =

4.82 3.82 3.82− −−2 3.8( ) 3.8( )

cosA 0.2022∠A 78°

2 B 180 78

2 B 102B 51

C 51

∠ = ° − °∠ = °∠ = °

∠ = °

b)

s =4.8+ 2 3.8( )

2= 6.2

A = 2 s s − a( ) s − b( ) s − c( )= 2

6.2( )6.2 − 4.8 6.2 − 3.8( ) 6.2 − 3.8( )14

The area of the flower beds is approximately 14 m2. Chapter 8 Section 3 Question 9 Page 419

cosA =a2 − b2 − c2

−2bc

cosA =

212 242 172− −−2 24( )17( )

cosA 0.5196∠A 59°

The ship should sail at an angle of 90° – 59°, or 31° to the western shoreline.



a)

cosA =a2 − b2 − c2

−2bc

cosA =

6.02 4.1 5.0− 2 − 2

−2 4.1( ) 5.0( )cosA 0.1417∠A 82°

The pole makes an angle of approximately 82° with the ground.

b)

cos B =b2 − a2 − c2

−2ac

cos B =

4.12 6.0 5.0− 2 − 2

−2 6.0( ) 5.0( )cos B = 0.7365∠B 43°

− −−2 11.3( ) 8.9( )

The beam makes an angle of approximately 43° with the ground.

c) C 180 43 82

55∠ = ° − ° − °

= °

The angle between the pole and the beam is approximately 55°. Chapter 8 Section 3 Question 11 Page 419

11.72 11.32 8.92

cos B =

cos B 0.3481∠B 69.6°

°

A 69.6

C 180 69.6110.4

D 110.4

∠ = °

∠ = ° − °= °

∠ = °

The interior angles of the trapezoid are 69. , to the specified rounding. 6 ,110.4 , 69.6 , and 11 . 0 4° ° °


Chapter 8 Section 3 Question 12 Page 419 Answers will vary. Chapter 8 Section 3 Question 13 Page 419

cos A =a2 − b2 − c2

−2bc

cos A =

1002 80 80− 2 − 2

−2 80( ) 80( )cos A 0.2188∠A 77.4°

°

2 B 180 77.42 B 102.6

B 51.3

C 51.3

∠ = ° − °∠ = °∠ = °

∠ = °

The angles in the triangle are , to the specified rounding. There is only one possible solution with the given sides.

51.3 , 51.3 , and 77.4° °

Chapter 8 Section 3 Question 14 Page 419 Solutions for Achievement Checks are shown in the Teacher Resource.



a)

cos A =a2 − b2 − c2

−2bc

=a2 − b2 − b2

−2bb

=a2 − 2b2

−2b2

=a2

−2b2 +1

= 1− a2

2b2

b)

cos A = 1− a2

2b2

cos A = 1−

0.82

2 1.5( )2cos A 0.8578∠A 30.9°

°

2 B 180 30.9

2 B 149.1B 74.55

C 74.55

∠ = ° − °∠ = °∠ = °

∠ = °

The angles in the triangle are 74 , to the specified rounding. .55 , 74.55 , and 30.9° ° Chapter 8 Section 3 Question 16 Page 419

cos60° = a2 − a2 − a2

−2aa

=−a2

−2a2

=12


Chapter 8 Section 4 Solve Problems Using Trigonometry Chapter 8 Section 4 Question 1 Page 427 a) Two sides and the contained angle are known. Use the cosine law. b) Two sides and another angle are known. Use the sine law. c) The triangle is a right triangle. Use primary trigonometric ratios. d) Three sides are known. Use the cosine law. Chapter 8 Section 4 Question 2 Page 427 a) Answers will vary. An example is given below.

( )( )( )( )( )1 5.1 66°

2 2 2

2 2 2

4.2cos35AC

ACcos35 4.24.2AC

cos35AC 5.1

ACD 180 57 57ACD 66

AC DC 2 AC DC cosA

5.1 5.1 5

CD

.2 cos5.6

x

xx

° =

° =

=°

∠ = ° − ° − °∠ = °

= + −

= + −

The length of x is approximately 5.6 m.

b) Answers will vary.



( )( )( )( )( )5.9 9 6.7 40°

2 2 2

2 2 2

4.5sin50AC

ACsin50 4.54.5AC

sin50AC 5.9

ACD 90 504

6.7 5.

0

AC DC 2 AC DC cosACD

2 cos44.

x

xx

° =

° =

=°

∠ = ° − °= °

=

+ −

+ −

=

The length of x is approximately 4.4 cm.

b) Answers will vary. Chapter 8 Section 4 Question 4 Page 428

tan68° = 1.5x

x tan68° = 1.5

x = 1.5tan68°

x 0.6

tan56° = 1.5y

y tan56° = 1.5

y = 1.5tan56°

y 1.0

The width of the crater is approximately 1.0 + 0.6, or 1.6 km.


Chapter 8 Section 4 Question 5 Page 428 a) b)

sin M sin681 1.5

sin68sin M1.5

M 38

S 180 68 3874

1.5sin74 sin68

1.5sin74sin68

1.6

x

x

x

°=

°=

∠ °

∠ = ° − ° − °= °

=° °

°=

°

At this point, Mars is approximately 1.6 A.U. from the Earth. In kilometres, Mars is approximately 1.6 × 149 600 000, or 239 360 000 km from the Earth. c) Answers may vary. For example: The distance between the Earth and Mars is not always the

same. Both planets move around the Sun in different orbits at different speeds.


Chapter 8 Section 4 Question 6 Page 428 a) Let x represent the distance between Riverside and Humberville.

x2 = 122 172+ − 2 12( )17( ) cos74°( )x 18

The total length of the race is approximately 12 + 17 + 18, or 47 km.

b) sinR sin7417 18

17sin74sin R18

R 65

°=

°=

∠ °

H 180 74 65

41∠ = ° − ° − °

= °

The angles between the towns are 65°, 41°, and 74°, respectively. ∠R,∠H, and ∠D



2 2 2

1.5tan146

B 90 684

180 36 8460

1.5 1414.1

sin A sinC

sin s14.1

36 60°° insin60 14.1sin36

14.1sin36sin60

9.6

x

x

C

cc

a c

a

a

a

a

=

°

∠ = ° − °= °

∠ = ° − ° − °= °

= +

=

=

° = °°

=°

The height of the building is approximately 9.6 m. Chapter 8 Section 4 Question 8 Page 428

Biff can run to the eucalyptus in

181.5

= 12 s.

Find the distance from Rocco to the eucalyptus.

( )( )( )2 2 215 18 15 18 452 cos12.9

xx= + − °

It will take Rocco approximately

12.91.0

, or 12.9 s to reach the eucalyptus. Assuming there are no

obstacles in either path, Biff can beat him to it.



( )( )( )6 5.7 20°

2 2 2

2 2 2

1.9sin 20

1.9sin 205.6

90 2070

90 7020

2 cos2 cos

2.5.6 5.7 5.

0

n

n

n

xx

yy

v n m nm yvv

° =

=°

= ° − °= °

= ° − °= °

= + −

= + −

tan33° = w5.7

w 3.7

cos33° = 5.7k

k = 5.7cos33°

k 6.8

q = 90° − 33°q = 57°

r = 90° − 57°r = 33°

s = 180° − 33° − 86°s = 61°

6.8sin33 sin86

6.8sin33sin86

3.7

u

u

u

=° °

°=

°

6.8

sin61 sin866.8sin61

sin866.0

t

t

t

=° °

°=

°

Assuming that the bridge is symmetric, the length of materials needed is approximately

2 5.6 + 5.6 + 3.8+ 2.0 + 5.7 + 3.7 + 6.8+ 3.7( )+ 6.0, or 79.8 m.



cos20° = x2.0

x = 2.0cos20°x 1.9

sin 20° = h2.0

h = 2.0sin 20°h 0.68

0.68sin15

0.68sin152.6

q

q

q

° =

=°

0.68tan15

0.68tan152.5

y

y

y

° =

=°

( )( )( )2 2 21.9 2.5 1.2 cos

1.8zz= + − 9 2.5 45°

Assuming the paths are straight, they hiked a total of approximately 2.0 + 2.6 +1.8, or 6.4 km. Chapter 8 Section 4 Question 11 Page 429

102 = s2 + 52

s 8.66

tan30° = x5

x = 5tan30°x 2.89

2 2 2

2 2

274.9956

2

8.35218

2.

.

8

2

9s x h

hh

h

= +

= +

8.66= +

The height of the tetrahedron is approximately 8.2 cm.



a) 6400tan0.0025

6400tan0.0025146 677 196

d

d

d

° =

=°

Then, the distance from Earth to the Sun is about, 146 677 196 – 6400, or 146 670 796 km, which is close to 149 600 000 km, as given in question 5. b) Answers will vary. Chapter 8 Section 4 Question 13 Page 429

a)

sinθ90

=sin60°

100100sinθ = 90sin60°

sinθ =90sin60°

100sinθ 0.7794

θ 51°

You must aim at approximately S51ºE.

b) Let X represent the angle between the heading and the wind.

X = 180° − 60° − 51°= 69°

gssin69°

=100

sin60°gssin60° = 100sin69°

gs =100sin69°

sin60°gs 108

The ground speed is approximately 108 km/h.


Chapter 8 Section 4 Question 14 Page 429 a) Assume that the angle up to Javier's balcony is half

the angle down to Raquel's balcony.

tan12° = x30

x = 30 tan12°

tan 24° = 2x30

2x = 30 tan 24°

x + 2x = 30 tan12° + 30 tan 24°19.7

Raquel and Javier live about 19.7 m apart.

b) Answers will vary. For example: Assume that the angle up to Javier's balcony is half the angle

down to Raquel's balcony. Draw a horizontal line through H. Use the tangent ratio in the two resulting right triangles.

Chapter 8 Section 4 Question 15 Page 429 a) The longest rod fits from the bottom front left corner to the top back right corner (or similarly).

The length of the rod, l, is the hypotenuse of the right triangle, whose legs are the diagonal of a rectangular face of the prism, and the height of a square base of the prism.

Find the diagonal of a rectangular face. Find the length of the rod.

d 2 = w2 + 2w( )2d 2 = 5w2

d = 5w

l2 = d 2 + w2

l2 = 5w( )2 + w2

l2 = 5w2 + w2

l2 = 6w2

l = 6w

b)

cosθ = w6w

θ 65.9°

cosθ = 2w6w

θ 35.3°

The rod makes an angle of approximately 65.9° with the base edge. The rod makes an angle of approximately 35.3° with the side edge.



( )( )( )

( )( )

2 2 2

2 280100 117

°

2

100 80 100 80 80

100 117

2 cos117

cos2

cos 0.738842

xx

y

yy

= + −

− −=

−

°

60° + 42° = 102° The ship must travel for approximately 117 km at a bearing of approximately N102ºE, or S78ºE. Chapter 8 Section 4 Question 17 Page 429 Answers will vary.


Chapter 8 Review Chapter 8 Review Question 1 Page 430

xsin X

=y

sin Yx

sin

=sin

1270°55°

x sin70° = 12sin55°

x =12sin55°

sin70°x 10

The length of x is approximately 10 cm. Chapter 8 Review Question 2 Page 430

sin Pp

=sinQ

qsin P

=sin62°

5.23.95.2sin P = 3.9sin62°

sin P =3.9sin62°

5.2sin P = 0.6622∠P 41°


Chapter 8 Review Question 3 Page 430

a)

sin Tt

=sin M

msin T

=sin51°

2824

sin T =24sin51°

28sin T 0.6661∠T 42°

K 180 42 5187

∠ = ° − ° − °= °

sin K sin M

sin sin28sin87

sin

2887 51° °

5136

k m

k

k

k

=

=

°=

°

In ΔKMT, is 42°, ∠T ∠K is 87°, and k is 36 km, to the specified rounding.

b)

∠C = 180° − 47° − 80°= 53°

nsin N

=r

sin Rn

sin

=sin

2780°47°

n =27sin 47°

sin80°n 20

csin N

=r

sin Rc

sin=

sin2780°53°

c =27sin53°

sin80°c 22

In ΔNRC, is 53°, n is 20 mm, and c is 22 mm, to the specified rounding. ∠C


Chapter 8 Review Question 4 Page 430 a)

sin B sin R

sin B sin

62sin B 55sin8255sin82sin B

62sin B 0.8785

B 61

W 180 82 6137

sin W sin R

sin sin

8255 62

6237 82

°

° °sin82 62sin37

62sin37sin82

38

b r

w r

w

w

w

w

=

=

= °°

=

∠ °

∠ = ° − ° − °= °

=

=

° = °°

=°

The red ball is approximately 38 cm from the black ball.

b) The measure of is 37° and the measure of ∠W ∠B is 61°, to the specified rounding. Chapter 8 Review Question 5 Page 430 t2 = a2 + v2 − 2av cosT( )t2 =

162 212+ − 2 16( ) 21( ) cos63°( )

t 20 The length of t is approximately 20 m.



( )( )( )( )

2 2 2

2 2 267 85 67 85 39

3985 54

°

°

2 cos U

2 cos54

sin P sin U

sin P sin

54sin P 85sin3985sin39sin P

54sin P 0.9906

P 82

D 180 82 3959

u d p dp

uu

p u

= + −

= + −

=

=

= °°

=

∠ °

∠ = ° − ° −°

°=

In ΔDPU, is 82°, ∠ is 59°, and u is 54 mm, to the specified rounding. ∠P D

b)

( )( )( )( )

2 2 2

2 2 211 14 11 14 77

7714 16

°

°

2 cosW

2 cos16

sinQ sin W

sinQ sin

sinQ 14sin7714sin77sinQ

16sinQ 0.8526

Q 58

E 180 58 7745

w e q eq

ww

q w

= + −

= + −

=

=

= °°

=

∠ °

∠ = ° − ° −°

°=

16

In ΔEQW, is 45°, is 58°, and w is 16 km, to the specified rounding. ∠E ∠Q



l2 = m2 + s2 − 2ms cos L( )l2 = 422 352+ − 2 42( ) 35( ) cos49°( )

l 33

The Mother Ship and Space Station are approximately 33 km apart.

b) sinS sin L

sinS sin

33sinS 35sin 4935sin 49sinS

33sin

4935 33

°

S 0.8004S 53

s l=

=

= °°

=

∠ °

From the Space Station, the shuttle and Mother Ship appear to be separated by approximately 53°.


cos B =b2 − a2 − c2

−2ac

cos B =

262 322 412− −−2 32( ) 41( )

cos B 0.7732∠B 39°



a)

cosS =s2 − f 2 − v2

−2 fv

cosS =

3.52 3.02 2.9− − 2

−2 3.0( ) 2.9( )cosS 0.2966∠S 72.7°

cos V =v2 − f 2 − s2

−2 fs

cos V =2.92 3.02 3.52− −−2 3.0( ) 3.5( )

∠F 180° − 52.3° − 72.7°55.0°

cos V 0.6114∠V 52.3°

In ΔVSF, is 52.3°, is 55.0°, and ∠V ∠F ∠S is 72.7°, to the specified rounding.

b)

( )( )

( )( )

2 2 2

2 2 2

2 2 2

2 2 2

cosS2

cosS2

cosS 0.4560S 62.9

cosZ2

cosZ2

1921 13

1321 19

21 13

21 19

cosZ 0.7932Z 37.5

B 180 37.5 62.979.6

s b zbz

z b sbs

− −=

−− −

=−

∠ °

− −=

−− −

=−

∠ °

∠ = ° − ° − °= °

In ΔSBZ, is 79.6°, is 37.5°, and ∠B ∠Z ∠S is 62.9°, to the specified rounding.



cos W =w2 − m2 − h2

−2mh

cos W =

5.62 4.82 7.02− −−2 4.8( ) 7.0( )

cos W 0.6054∠W 53°

The angle from the pilot to the water tower is approximately 90° – 53°, or 37°.

cos M =m2 − w2 − h2

−2wh

cos M =

4.82 5.62 7.02− −−2 5.6( ) 7.0( )

cos M 0.7311∠M 43°

The angle from the pilot to the monument is approximately 90° – 43°, or 47°. Chapter 8 Review Question 11 Page 431 a)

b)

sin30° = h2.0

h = 2.0sin30°h = 1

cos30° = x2.0

x = 2.0cos30°x 1.7

3.52 = 12 + y2

y 3.4

The distance between the helicopters is approximately 3.4 + 1.7, or 5.1 km. c) The altitude of the helicopters is 1 km.



D 180 54 5175

sin A sin D

sin sin100

51 75=

° °sin75 100sin51

100sin51sin75

80

tan 418080tan 4170

a d

a

a

a

a

h

hh

∠ = ° − ° − °= °

=

° = °°

=°

° =

= °

The height of Percé Rock is approximately 70 m.


Chapter 8 Practice Test Chapter 8 Practice Test Question 1 Page 432

bsin B

=a

sin Ab

sin

=sin

2574°45°

bsin74° = 25sin 45°

b25sin 45°

sin74°b 18

+ − 2 4.3( ) 3.6( ) cos(

In ΔABC, b is approximately 18 cm. Chapter 8 Practice Test Question 2 Page 432

r 2 = k 2 + t2 − 2kt cos R( )r 2 = 4.32 3.62 36°)

r 2.5 In ΔKRT, r is approximately 2.5 m. Chapter 8 Practice Test Question 3 Page 432

sin Yy

=sin N

nsin Y

=sin65°

161316sin Y = 13sin65°

sin Y =13sin65°

16sin Y 0.7364∠Y 47°

Chapter 8 Practice Test Question 4 Page 432

cos H =h2 − e2 − s2

−2es

cos H =

212 242 182− −−2 24( )18( )

cos H 0.5313∠H 58°


Chapter 8 Practice Test Question 5 Page 432 a)

x2 = 2.22 2.82 80°+ − 2 2.2( ) 2.8( ) cos( ) x 3.2

The wires are approximately 3.2 m apart.

b) sin A sin802.8 3.2

2.8sin80sin A3.2

A 60

C 180 60 8040

°=

°=

∠ °

∠ = ° − ° − °= °

Assuming the ground between the wires is flat, the left wire makes an angle of approximately 60° with the ground, and the right wire makes an angle of approximately 40° with the ground.


cos A =a2 − b2 − c2

−2bc

cos A =202 352 352− −−2 35( ) 35( )

cos A 0.8367∠A 33.2°

Cheryl must hit within an angle of approximately 33.2°.



( )( )( )( )

2 2 2

2 2 27.2 8.3 7.2 8.3 68

2 cosX

2 cos8.7

sin Y sin X

sin Y sin

8.7sin Y 7.2sin687.2sin68sin Y

8.7sin Y 0.7673

Y 50

Z 180 852

0 66

x y z yz

xx

y x

= + −

= + −

=

=

= °°

=

∠ °

∠ = ° − ° − °= °

687.2 8 7.

°

°

In ΔXYZ, is 62°, is 50°, and x is 8.7 m, to the specified rounding. ∠Z ∠Y


Chapter 8 Practice Test Question 8 Page 432 a)

b)

sin Ww

=sin N

nsin W

=sin54°

2.31.82.3sin W = 1.8sin54°

sin W =1.8sin54°

2.3sin W 0.6331∠W 39°

T 180 39 54

87∠ = °− ° − °

= °

sinT sin N

sin sin2.3

87 54° °sin54 2.3sin87

2.3sin87sin54

2.8

t n

t

t

t

t

=

=

° = °°

=°

In ΔNTW, is 39°, is 87°, and t is 2.8 km, to the specified rounding. ∠W ∠T



( )( )( )

( )( )(

2 2

2 2 2

2 c s5.8

sin sin

5.8sin 6.2sin 466.2sin 46sin

5.8sin 0.7

8 6.2

5.8 5.8 5.8 5.8 80

68950

2 cos7.5

xx

z

z

z

zz

yy

= +

=

= °°

=

∠ °

= + − )

2 o8 6.2 46

466.2 5.8

− °

°

°

Assuming the two blue triangles are congruent, the length of trim needed is approximately

or 47.5 cm. 16 + 2 × 6.2 + 2× 5.8+ 7.5, Chapter 8 Practice Test Question 10 Page 432

AXB 180 60 8238

AX ABsin ABX sin AXB

AXsin sin

5082 38

=° °

AXsin38 50sin8250sin82AX

sin38AX 80

tan 438080 tan 4375

h

hh

∠ = ° − ° − °= °

=

° = °°

=°

° =

= °

The height of the Peace Tower is approximately 75 m.



∠P = 180° − 65° − 48°= 67°

ssinS

=p

sin Ps

sin=

sin5067°48°

ssin67° = 50sin 48°

s =50sin 48°

sin67°s 40

The plane was approximately 40 km from Hamilton. Chapter 8 Practice Test Question 12 Page 433

( )( )( )( )

2 2 2

2 2 2400 300 400 300 60

2 cosT

2 c361

sin B sinT

sin B sin

361sin B 400sin60400sin60sin B

361sin B 0.9596

B 74

t b f bf

tt

b t

= + −

= + −

=

=

= °°

=

∠ °

60400 361

°

°

os

Felipe is approximately 361 m from the boat. To return, he must swim in the approximate direction S74ºW.



a)

tan60° = a20

a = 20 tan60°a 34.6

tan50° = b20

b = 20 tan50°b 23.8

The height of the blue jay tree is approximately 1

.5 + 34.6, or 36.1 m.

The height of the cardinal tree is approximately 1.5 + 23.8, or 25.3 m.

) The blue jay is approximately 36.1 – 25.3, or 10.8 m higher than the cardinal.

2

The birds are approximately 41.4 m apart.

b

2 210.8 4041.4

dd= +



A =12

bh

=12

126 b 12( )b = 21

J 180 53 8245

sin J sin W

sin

∠

sin21

45 82sin82 21sin 45

21sin 45sin82

15

j w

j

j

j

j

= ° − ° − °= °

=

° = °°

=°

=° °

sinSs

=sin W

wsinS

=sin82°

21

1721sinS = 17 sin82°

sinS =17sin82°

21sinS 0.8016∠S 53°

The perimeter is approximately 15 + 21 + 17, or 53 km.

5380

= 0.6625

At 80 km/h, the time taken to drive the perimeter is approximately 0.6625 h, or 40 min.



p2 = a2 + j2 − 2aj cos P( )p2 = 122 162+ − 2 12( )16( ) cos52°( )p 13

sin38° = h16

h = 16sin38°h 10

The altitude of the jet is approximately 13 km. The altitude of the plane is approximately 10 km. Chapter 8 Practice Test Question 16 Page 433 Solutions for Achievement Checks are shown in the Teacher Resource.


Chapters 7 and 8 Review Chapters 7 and 8 Review Question 1 Page 434 ∠B = ∠E (right angles) ∠ACB = ∠DCE (opposite angles) ∠A = D (angle sum of a triangle is 180°) ∠ Therefore, ΔABC ~ ΔDEC because corresponding pairs of angles are equal. Chapters 7 and 8 Review Question 2 Page 434

h7=

2613

13h = 182h = 14

q28

=1326

26q = 364q = 14

h = 14 cm, and q = 14 cm.


Chapters 7 and 8 Review Question 3 Page 434

a) 6.4tan E4.8

E 53

C 90 5337

=

∠ °

∠ = ° − °= °

b) 31sin Y53

Y 36

X 90 3654

=

∠ °

∠ = ° − °= °

c) 8.0sinT9.5

T 57

U 90 5733

=

∠ °

∠ = ° − °= °

d) 16tan M23

M 35

N 90 3555

=

∠ °

∠ = ° − °= °



a)

tan56° = 9x

x tan56° = 9

x = 9tan56°

x 6.1


b)

sin 65° = x33

x = 33sin 65°x 29.9

The length of x is approximately 29.9 m.

c)

cos52° = x8.8

x = 8.8cos52°x 5.4

The length of x is approximately 5.4 km.

d)

sin31° = 51x

x sin31° = 51

x = 51sin31°

x 99.0



Chapters 7 and 8 Review Question 5 Page 434 a)

2 2

2

9.3tan A7.5

A 51

C 90 5139

9.3 7.5142.7411.9

bbb

=

∠ °

∠ = ° − °= °

= +

=

2

In ΔABC, is 51°, is 39°, and b is 11.9 cm, to the specified rounding. ∠A ∠C

b)

tan59° = f5.8

f = 5.8 tan59°f 9.7

∠H = 90° − 59°= 31°

sin31° = 5.8g

g sin31° = 5.8

g =5.8

sin31°g 11.3

In ΔGFH, is 31°, f is 9.7 m, and g is 11.3 m, to the specified rounding. ∠H


x1.5

=7.62.7

2.7x = 11.4x 4.2

The height of the support at A is approximately 4.2 m.



a)

tan12° = x96

x = 96 tan12°x 20

The height of the rock ledge is

approximately 20 m.

b)

tan 22° = y96

y = 96 tan 22°y 39

The height of the tree is approximately 20 + 39, or 59 m.


xsin X

=y

sin Yx

sin

=sin

1182°52°

x sin82° = 11sin52°

x =11sin52°

sin82°x 9

+ − 2 5.3( ) 4.6( ) cos(

In ΔXYZ, x is approximately 9 cm. Chapters 7 and 8 Review Question 9 Page 435

c2 = a2 + b2 − 2ab cosC( )c2 = 5.32 4.62 46°)

c 3.9 In ΔABC, c is approximately 3.9 cm.



sin Pp

=sinQ

qsin P

=sin63°

1.61.41.6sin P = 1.4sin63°

sin P =1.4sin63°

1.6sin P 0.7796∠P 51°


cos E =e2 − d 2 − f 2

−2df

cos E =

402 312 382− −−2 31( ) 38( )

cos E 0.3417∠E 70°



a)

sin Bb

=sinC

csin B

=sin 44°

131713sin B = 17sin 44°

sin B =17sin 44°

13sin B 0.9084∠B 65°

A 180 65 4471

∠ = ° − ° − °= °

sin A sin B

sin sin17

71 65=

° °sin65 17sin71

17sin71sin65

18

a b

a

a

a

a

=

° = °°

=°

In ΔABC, a is 18 mm, is 65°, and ∠B ∠A is 71°, to the specified rounding.

b)

( )( )( )( )

2 2 2

2 2 2

2 cos R

2 cos26

sinS sin5431 26

31sin54sinS26

S

31 2

75

T 180 75 545

3

1

4 1

r s t st

rr

= + −

= + −

°=

°=

∠ °

∠ = ° − ° − °= °

24 54°

In ΔRST, r is 26 m, is 75°, and ∠S ∠T is 51°, to the specified rounding.


c)

∠Y = 180° − 81° − 32°= 67°

xsin X

=w

sin Wx

sin

ysin Y

=w

sin Wy

sin

81°=

sin1632°

=sin

1632°

x sin32° = 16sin81°

x =16sin81°

sin32°x 30

67°y sin32° = 16sin67°

y =16sin67°

sin32°y 28

In ΔWXY, x is 30 cm, y is 28 cm, and ∠Y is 67°, to the specified rounding.

d)

( )( )( )( )

( () )

2 2 2

2 2 2

2 2 2

2

2 cos E

2 cos36

cos F2

32 21 3

cos F2

cos F 0.4716F

21 3

62

G 18

6

0 62 8434

e f g fg

ee

f g ege

= + −

= + −

− −=

−

− −=

−

∠ °

∠ = ° − ° − °= °

2 2

2 21 84

3221 36

°

In ΔEFG, e is 36 km, is 62°, and ∠F ∠G is 34°, to the specified rounding.



a)

( )( )

2 2 2

2 2 28.7 9.6

cosA2

cosA2

cosA 0.7280A 43.3

C 180 61.3 43.375.4

a b cbc

− −=

−− −

=−

∠ °

∠ = ° − −= °

cos B =b2 − a2 − c2

−2ac

cos B =

8.72 6.82 9.6− − 2

−2 6.8( ) 9.6( )6.8

8.7 9.6

cos B 0.4803∠B 61.3°

In ΔABC, is 43.3°, is 61.3°, and ∠A ∠B ∠C is 75.4°, to the specified rounding.

b)

( )( )

2 2 2

2 2 211.4 12.5

cosT2

cosT2

cosT 0.632T 50.8

U 180 70.1 50.859.1

t u vuv

− −=

−− −

=−

∠ °

∠ = ° − −= °

cos V =v2 − u2 − t2

−2vt

cos V =

12.52 11.42 10.32− −−2 11.4( )10.3( )

10.311.4 12.5

cos V 0.3398∠V 70.1°

In ΔTUV, is 70.1°, is 59.1°, and ∠V ∠U ∠T is 50.8°, to the specified rounding.


∠C = 180° − 43° − 55°= 82°

asin A

=b

sin Ba

sin=

sin6082°43°

a sin82° = 60sin 43°

a =60sin 43°

sin82°a 41

The ship is approximately 41 km from the second observation point.



a2 = 17302 19802 36°+ − 2 1730( )1980( ) cos( ) a 1171

The length of the tunnel is approximately 1171 m. Chapters 7 and 8 Review Question 16 Page 435

BDA 180 53 4285

BD ABsin BAD sin ADB

BDsin sin

BDsin85 6000sin 426000

600042 58 °°

sin 42BDsin85

BD 4030

tan3840304030tan3813 49

h

hh

∠ = ° − ° − °= °

=

=

° = °°

° =

°

=°

=

The height of the mountain is approximately 3149 m.


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