Chapter 12 Exercise 12.1
Q. 4.
(i) x2 = 42 + 32
x=5
Q. 1.
(i)
x2
=
212
+
202
y2 + 52 = 132
x = 29
y2 = 132 − 52
(ii) x2 = 132 + 842
y2 = 144
x2 = 7225
y = 12
x = 85
(ii)
132
+ 842 = x2
(iii) x2 + 42 = 52
x2
x = 85
=9
852 = 362 + y2
x=3
y2 = 5929
(iv) x2 + 102 = 262
x2
y = 77
= 576
(iii) 1002 = 282 + x2
x = 24
(v)
252
=
x2
+
x = 96
72
800 − 282 = y2
x2 = 576
16 = y2
x = 24
(vi)
412
−
92
y=4
=
x2
(iv)
x = 40
Q. 2.
x2
+ 62
x2 = 100
x = 10
(i) x2 = 32 + 12
262 = 102 + y2
x2 = 10
___
y2 = 576
x = √10
y = 24
(ii) x2 = 12 + 12
x2 = 2
=
82
Q. 5.
__
80
x = √2
(iii) x2 = 32 + 52
60
x2 = 34
___
x = √34
(i) 280 − 2(80) = 120
(iv) 72 = 52 + x2
120 ÷ 2 = 60 cm
x2 = 24
___
(ii)
x = √24
x2
= 802 + 602
x2 = 10000
Q. 3.
x = 100
6.52 = 62 + x2
6.5
6
(iii) Area = 80 × 60 = 4800 cm2
x2 = 6.25
x = 2.5 m
x
Active Maths 3 Book 2 – Strands 1–5 – Ch 12 Solutions
1
Q. 6. (a)
Side a
Side b
3
4
5
6
8
10
9
12
15
12
16
20
15
20
25
18
24
30
21
28
35
a2
b2
c2
(b)
Hypotenuse c
Tick if true
9
16
25
ü
36
64
100
ü
81
144
225
ü
144
256
400
ü
225
400
625
ü
324
576
900
ü
441
784
1225
ü
Q. 7. The width of a door frame is 84 cm and its height is 187 cm
x2 = 842 + 1872
84 cm
x2 = 42025
______
x = √ 42025
187 cm
X
Q. 8. (a)
2
x = 205
Side a
Side b
Hypotenuse c
2000
2100
2900
200
210
290
20
21
29
2
2.1
2.9
0.2
0.21
0.29
0.02
0.021
0.029
0.002
0.0021
0.0029
Active Maths 3 Book 2 – Strands 1–5 – Ch 12 Solutions
(b)
a2
b2
c2
Tick if true
4000000
4410000
8410000
ü
40000
44100
84100
ü
400
441
841
ü
4
4.41
8.41
ü
0.04
0.0441
0.0841
ü
0.0004
0.000441
0.000841
ü
0.000004
0.00000441
0.00000841
ü
Q. 9. Triangle with sides: 85, 77, 36
852 = 7225, 772 + 362 = 5929 + 1296 = 7225
⇒ Triangle is right-angled.
Q. 10. Triangle with sides: 7, 24, 25
252 = 625, 72 + 242 = 49 + 576 = 625
⇒ Triangle is right-angled.
Q. 11. Triangle with sides 11, 60, 61
612 = 3721, 112 + 602
= 121 + 3600 = 3721
⇒ Triangle is right-angled.
Exercise 12.2
Q. 1.
Q. 2.
Q. 3.
5
5
12
(i) sin A = ___ cos A = ___ tan A = ___
13
13
12
20
21
21
(ii) sin A = ___ cos A = ___ tan A = ___
20
29
29
40
40
40
42
42
42
(i) sin A = ___ cos A = ___ tan A = ___ sin B = ___ cos B = ___ tan B = ___
58
58
42
58
58
40
3
3
3
2
2
2
___
___
___
___
(ii) sin A = _____
cos A = _____
tan A = __ sin B = _____
cos B = _____
tan B = __
2
3
13
√
√ 13
√ 13
√ 13
2
4
2
4
2
4
___
___
___
___
(iii) sin A = _____
cos A = _____
tan A = __ sin B = _____
cos B = _____
tan B = __
4
2
√ 20
√ 20
√ 20
√ 20
(i) 0.2588
(x) 0.4791
(ii) 0.8660
(xi) 0.7218
(iii) 3.7321
(xii) 0.9823
(iv) 0.2419
(xiii) 0.5210
(v) 0.9004
(xiv) 0.2830
(vi) 0.0872
(xv) 0.8581
(vii) 0.2126
(xvi) 0.5490
(viii) 0.5
Q. 4.
(xvii) 0.8934
(ix) 5.6713
(xviii) 7.4947
(i) 38.26
(vi) 82.84
(ii) 29.61
(vii) 58.54
(iii) 19.76
(viii) 46.23
(iv) 25.94
(ix) 64.93
(v) 20.62
(x) 63.11
Active Maths 3 Book 2 – Strands 1–5 – Ch 12 Solutions
3
Q. 5.
Q. 6.
Q. 7.
(i) 2° 30ʹ
(iv) 25° 24ʹ
(ii) 2° 15ʹ
(v) 1° 12ʹ
= 4.37
(iii) 2° 45ʹ
(vi) 0° 20ʹ
(ii) y = 50 sin 29
(i) 2.52°
(iv) 70.37°
(ii) 10.67°
(v) 11.62°
(iii) 25.83°
(vi) 33.55°
(i) 75° 57ʹ
(iv) 41° 34ʹ
(ii) 48° 14ʹ
(v) 17° 42ʹ
(iii) 43° 55ʹ
(vi) 26° 49ʹ
(vii) 73° 54ʹ
(ix) 22° 12ʹ
(viii) 13° 57ʹ
(x) 58° 01ʹ
Q. 8. tan B =
Q. 2.
= 24.24
(iii) y = 18 cos 35
= 14.74
(iv) y = 12 tan 30°
= 6.93
(v) y = 26 cos 50
= 16.71
(vi) y = 10 sin 30
=5
9
___
40
B=
tan−1
( 40 )
9
___
Q. 3.
= 64.28
y = 64.28 tan 10°
≅ 13°
y = 11.33
Q. 4.
≅ 3.6
______
y = √ 21.96 ≅ 4.7
Q. 5.
x=5
x
___
= cos 55
18
x = 10.32
x
___
(iv)
= sin 35°
20
x = 11.47
(v) x = 16 sin 30°
x=8
x
(vi) ___ = cos 45°
40
x = 40 cos 45°
x = 28.28
4
Active Maths 3 Book 2 – Strands 1–5 – Ch 12 Solutions
z2 = 132 − 4.72
z2 = 146.91
z ≅ 12
y2 = 3.62 + 32
Exercise 12.3
x
(i) __ = tan 60°
5
x ≅ 8.66
x
___
= cos 60°
(ii)
10
x = 10 cos 60°
x = 3 tan 50°
= 3.5753
≅ 28° 04ʹ
(iii)
x = 100 sin 40°
= 12.6804
15
Q. 9. cos C = ___
17
= 28.0725
Q. 1.
(i) y = 7 tan 32
4
(i) sin A = __
7
4
A = sin−1 __
7
A = 34.85°
A = 35°
2
(ii) cos A = __
6
2
A = cos−1 __
6
A = 70.52°
A = 71°
5
(iii) tan A = __
6
5
A = tan−1 __
6
A = 39.8°
A = 40°
5
(iv) tan A = __
3
5
A = tan−1 __
3
A = 59.03°
200
____
x = tan 30°
200 = x tan 30°
200
x = _______
tan 30°
x = 346 m
Q. 3.
A = 59°
7
(v) sin A = __
8
7
A = sin−1 __
8
A = 61.04°
x
(i) sin 47° = ___
22
x = 22 sin 47
Q. 4.
(ii) x = sin 63°
16
x = ______
sin 63°
x = 18 m
A = 61°
4
(vi) tan A = __
4
A = tan−1 1
A = 45°
x
tan 69 = ______
202.4
x = 202.4 tan 69
Q. 5.
Exercise 12.4
Q. 1.
Q. 2.
≅ 16 metres
16
___
x = 527.3 m
x
(i) sin 67° = ___
46
x = 46 sin 67°
x ≅ 42 km
46
(ii) ___ = 23 km/hr
2
42
(iii) ___ = 21 km/hr
2
x
________
= tan 77°
103.7 m
x = 103.7 tan 77
x
tan 33° = __
5
x = 5 tan 33°
Q. 6.
= 3.25
Liam’s height 1.7 m
Tree
1.7 + 3.25 = 4.95 metres
x
tan 78° = ____
100
x = 100 tan 78°
Q. 7.
= 470.5 m
= 471 m
x = 449.2 m
Q. 8.
D
55°
x
y
z
35°
6
find x
15
find y
6
__
(7.3246)2 = (6)2 + z2
cos 35 = x
cos 35 = .81915
6
.81915 = __
x
(.81915)(x) = 6
z2 = 17.6497
z = 4.2
(4.2)2
+ (15)2 = (y)2
(y)2 = 242.64
x = 7.3246 m = 7.32 m
Multiplied by 2
y = 15.5769 m = 15.58 m
Multiplied by 2
= 14.64 m
= 31.16
Total length of wire = (15.58 + 7.32) × 2
= 45.8 m
Active Maths 3 Book 2 – Strands 1–5 – Ch 12 Solutions
5
Q. 9.
h
Q. 10. (a) tan 52 = __
x
(b) ∠ADB = 90 − 35 = 55°
15
∠ADC = tan−1 ___
4.2
∠ADC = 74.36°
h
tan 40 = ______
x + 10
(b) h = x tan 52°
C
(c) h = (x + 10) tan 40
(d) x tan 52 = (x + 10) tan 40
y
x = 19.03m
6
5
(e) x tan 52 − x tan 40 = 10 tan 40
x (tan 52 − tan 40) = 10 tan 40
10 tan 40
x = ______________
tan 52 − tan 40
x = 19.03 m
√11
50°
x
A
B
D
(a) |BD|2 = 62 − 52
(f) h = (19.03)(tan 52°)
|BD|2 = 36 − 25
|BD|2
h = 24.36 m
= 11
___
|BD| = √ 11
|BD| = 3.3166 feet
3 feet 4 inches
5
(b) sin(∠CDB) = __
6
5
∠CDB = sin−1 __
6
∠CDB = 56.44°
Exercise 12.5
Q. 1.
5
tan 50 = __
x
Q. 2.
1
√3
(i) 30°
__ 2
(iii) 12 + ( √ 3 ) = |AC|2
1 + 3 = |AC|2
x = 4.2 feet
|AC| = √ 4
5
__
__
|AC| = 2
Q. 3.
y = 6.527
y = 6.53 feet
Total length = 4.2 + 6.53 + 6 + 3.3
= 20.03
+ 5% wastage
6
2__
3
√___
2
1__
___
√3
60°
|AB|
(ii) tan 60° = _____
1
__
√ 3 = |AB|
sin 50 = .766
5
__
y = .766
(.766)(y) = 5
100
1
__
45°
1__
___
√2
1__
___
√2
x = 4.195
sin 50 = y
20.03
______
sin A
tan A
tan 50 = 1.1917
5
__
x = 1.1917
(1.1917)(x) = 5
find y
30°
cos A
∠CDB = 56°
(c) find x
A
Q. 4.
× 105 = 21.03 feet
Active Maths 3 Book 2 – Strands 1–5 – Ch 12 Solutions
(i) 180 − 90 − 45 = 45°
|BC|
__
(ii) cos 45° = _____
7√ 2
|BC|
1__ _____
___
__
=
√ 2 7√__2
7√__2
|BC| = ____
√2
|BC| = 7
x __
(i) cos 45° = ____
4√ 2
x=4
Isoceles Δ so y = 4.
(ii) 90°
__
3
√___
2
1
__
2
__
Q. 5.
(√ ) ( )
(√ ) ( )
1__
___
2
2
__ 2
2
1__
1 1
+ ___
= __ + __ = 1
2 2
2
√
(ii)
5√ 3
2
__ 2
3
1
Q. 6. ___ + __ + ( √ 3 )
2
2
3 __
1
__
= + +3=4
4 4
__ 2
2
2
3
√
1
1
___
__
___
__
Q. 7.
+
+
2
2
√3
3
1
4
1
__
+ __ + __ = __
3 4 4 3
1
__
sin 30°
2__ ___
1
_______
___
=
Q. 8.
= __ = tan 30°
cos 30° ___
√3 √3
2
Area = 37.5
A
10
( ) () ( )
1
Area = __ ab sin C
2
__
1
37.5 = __( 5√ 3 )(10) sin A
2 __
√3
sin A = ___
2
__
3
√
___
A = sin−1
2
A = 60°
Q. 10.
Exercise 12.6
15
1
Q. 1. __(12)(15) sin 55° = 73.72 square units
2
1
Q. 2. __(24)(20) sin 30° = 120 sq units
2
1
__
Q. 3.
(14)(18) sin 40° = 80.99 sq units
2
1
Q. 4. __(25)(30) sin 25° = 158.48 sq units
2
1
__
(10)(10) sin 70° = 46.98 sq units
Q. 5.
2
Q. 6.
80
Area = 30
36°
x
1
Area = __ ab sin C
2
1
__
30 = (15)(x)(sin 36)
2
x = 6.805
x=7
Q. 11. (a)
A
60
B
X
12
50
50°
50°
75
1
__
2
(12)(12) sin 80° = 70.91 sq units
D
1
Q. 7. __(7)(8.4) sin 62° = 25.96 sq units
2
1
Q. 8. __(9)(18.4) sin 82° = 82.8 sq units
2
Q. 9. (i)
602 + 752 − 502
cos X = _______________
2(60)(75)
|∠ABD| = 41.65°
B
6
A
Y
75
Area = 12
50
8
1
Area = __ ab sin C
2
1
__
12 = (6)(8) sin A
2
1
__
sin A =
2
1
A = sin−1 __
2
A = 30°
D
60
C
752 + 502 − 602
cos Y = _______________
2(75)(50)
|∠DBC| = (52.89)°
(b) |∠DAB| = 180° − (41.65° + 52.89°)
= 85.46°
Active Maths 3 Book 2 – Strands 1–5 – Ch 12 Solutions
7
Q. 12.
Q. 7. 0
Y
Q. 8. 1
6
4
Q. 9. 0
Q. 10. 0
X
Z
(i) cos ∠XYZ =
Q. 11. 1
3
__
5
Q. 12. 0
X
Q. 13. 0
5
4
Y
Z
3
Exercise 12.8
(XZ)2 = 52 − 32
(XZ)2 = 16
(XZ) = 4
4
sin ∠XYZ = __
5
1
__
(ii) Area = ab sin C
2
4
1
__
Area = (6)(4) __
5
2
48
Area = ___
5
Area = 9.6 units2
()
180 – θ
180 + θ
θ
S
A
T
C
Q. 1. cos 135°
360 – θ
180 − 135° = 45°
−cos 45°
1__
= −___
√2
Exercise 12.7
Q. 2. sin 150°
90°
180 − 150° = 30°
= sin 30°
1
= __
2
180°
0°, 360°
Q. 3. cos 240°
−cos 60°
1
= −__
2
270°
Q. 1. 0
Q. 2. −1
Q. 3. 0
Q. 4. sin 330°
360° − 330° = 30°
−sin 30°
1
= −__
2
Q. 5. tan 210°
210° − 180° = 30°
Q. 4. 1
Q. 5. 0
Q. 6. −1
8
240° − 180° = 60°
Active Maths 3 Book 2 – Strands 1–5 – Ch 12 Solutions
tan 30°
1__
= ___
√3
Q. 6. cos 315°
360° − 315° = 45°
cos 45°
1__
= ___
√2
Q. 7.
Q. 12. 0.34
sin 120°
180° − 120° = 60°
Q. 13. −0.64
Q. 14. −0.09
Q. 15. 0.84
cos 210°
210° − 180° = 30°
−cos 30°
__
3
√
___
=−
2
Q. 9.
__
tan 60° = √3
Q. 11. −0.82
sin 60°
__
3
√
___
=
2
Q. 8.
Q. 10.
Q. 16. 0.82
Q. 17. 0.64
Q. 18. −0.64
tan 300°
360 − 300 = 60°
−tan 60°
Q. 19. −0.18
Q. 20. −0.36
__
= −√ 3
1
Q. 21. sin A = __
2
0 ≤ A ≤ 360°
30°
30°
30°
30°
A = 30°
OR
sin is positive
A = 180 − 30°
A = 150°
1__
Q. 22. cos A = ___
√2
45°
45°
45°
45°
A = 45°
0 ≤ A ≤ 360
OR
__
Q. 23. tan A = √ 3
60°
60°
60°
60°
A = 60°
S
A
T
C
cosine is positive
A = 360° − 45°
A = 315°
0 ≤ A ≤ 360
tan is positive
OR
A = 180° + 60°
A = 240°
Active
A
ti Maths
M th 3 B
Book
k 2 – St
Strands
d 1
1–5
5 – Ch 12 S
Solutions
l ti
9
1__
Q. 24. sin A = −___
√2
45°
45°
45°
45°
0 ≤ A ≤ 360
sin is negative
A = 225°
OR
A = 360° − 45°
A = 315°
1__
Q. 25. tan A = −___
√3
30°
30°
30°
30°
A = 150°
0 ≤ A ≤ 360
tan is negative
OR
A = 315°
__
√3
___
Q. 26. sin A = −
0 ≤ A ≤ 360°
2
60°
60°
60°
60°
A = 240°
A = 360° − 45°
OR
A = 360 − 60
sin is negative
A = 300°
Exercise 12.9
10
x
Q. 1. ______ = ______
sin 50° sin 40°
10 sin 50°
x = _________
sin 40°
x = 11.92
20
x
Q. 2. ______ = ______
sin 30° sin 40°
20 sin 30°
x = _________
sin 40°
x = 15.56
10
10
x
Q. 3. _______ = ______
sin 125° sin 25°
10 sin 125°
x = __________
sin 25°
x = 19.38
20
x
Q. 4. ______ = _______
sin 30° sin 100°
20 sin 30°
x = _________
sin 100°
x = 10.15
10
x
Q. 5. ______ = ______
sin 70° sin 60°
10 sin 70°
x = _________
sin 60°
x = 10.85
Active Maths 3 Book 2 – Strands 1–5 – Ch 12 Solutions
Q. 6.
1
(iv) Area = __ ab sin C
2
1
9.65 = __(8)(C)(sin68.44)
2
C = 2.59
C
15
25
sin A sin 44° 26ʹ
Q. 9. _____ = __________
16
14
16 sin 44° 26ʹ
_____________
sin A =
14
16 sin 44° 26ʹ
A = sin−1 _____________
14
A = 53°
Q. 10.
A
50°
sin A sin 50
(i) _____ = ______
25
15
sin A = .45962
(
A = sin−1 .45962
A = 27.36°
(ii) C = 180 − (50 + 27.36)
45°
C = 102.64
1
(iii) Area = __ ab sin C
2
1
Area = __(15)(25)(sin 102.64)
2
Area = 182.96 units2
= 183.0 units2
sin A sin 140°
Q. 7. _____ = _______
25
12
12 sin 140°
sin A = __________
25
12 sin 140°
A = sin−1 __________
25
A = 18°
(
)
X
7
A
111°
(i)
12
sin 45
sin
A
_____
______
7
=
12
__
7√ 2
____
sin A =
24
A = 24.36°
A = 24°
1
(ii) Area = __ ab sin C
2
1
Area = __(7)(12)(sin 111)
2
)
Area = 39.21 units2
Q. 8.
8
9.3
53°8⬘
sin A sin (53° 8ʹ)
(i) _____ = __________
8
9.3
sin A = .93
A = sin−1 .93
A = 68.44°
A = 68° 26ʹ
(ii) €180 = (68.44 + 53° 8ʹ)
C = 164.69°
1
(iii) Area = __ ab sin C
2
1
__
Area = (8)(9.3) (sin 164.69°)
2
Area = 9.65 units2
1
(iii) Area = __ ab sin C
2
1
39.21 = __(12)(x)(sin 24)
2
x = 16.07
|PQ|
80
Q. 11. ______ = ______
sin 60° sin 70°
80 sin 60°
|PQ| = _________
sin 70°
= 74 m
|RR|
80
______
= ______
sin 50° sin 70°
80 sin 50°
|PR| = _________
sin 70°
= 65 m
Active Maths 3 Book 2 – Strands 1–5 – Ch 12 Solutions
11
Q. 12.
Exercise 12.10
115°
120
80
40°
25°
Q. 1. a2 = 42 + 52 − 2(4)(5) cos 60°
a2 = 21
sin 40°
= ______
80
120
80 sin 40°
_________
sin x =
120
80 sin 40°
x = sin−1 _________
120
x = 25°
1
Area = __(80)(120) sin 115°
2
= 4350.28 metres2
___
sin x
____
(
Q. 13. (a)
)
Q. 2. a2 = 52 + 102 − 2(5)(10) cos 70°
a2 = 90.7980
a = 9.53
Q. 3. a2 = 92 + 122 − 2(9)(12) cos 80°
a2 = 187.4920
a = 13.69
Q. 4. a2 = 52 + 52 − 2(5)(5) cos 45° 34ʹ
65
50
a = √21 = 4.58
a2 = 14.9961
A
a = 3.87
65°
50°
Q. 5. a2 = 152 + 122 − 2(15)(12) cos 85°
50
sin 50
______
sin 65
= ______
50
A
|A| = 42.26 cm
a2 = 337.6239
a = 18.37
52 + 82 − 72
Q. 6. cos A = ____________
2(5)(8)
−1
A = cos (0.5)
(b) V = (.4)(.98)(.48)
V = .18816 m3
(c)
= 60°
98
x
48
A = 120°
x2 = (48)2 + (98)2
x2
52 + 32 − 72
Q. 7. cos A = ____________
2(5)(3)
−1
A = cos (−0.5)
52 + 62 − 42
Q. 8. cos A = ____________
2(5)(6)
−1
A = cos (0.75)
= 11,908
x = 109.12 cm
The diagonal is greater than
the length of the bar so the
bar will lie flat on the floor
of the box.
A = 41°
102 + 92 − 32
Q. 9. cos A = _____________
2(10)(9)
−1
A = cos (0.9556)
A = 17°
(d)
40
109.12
x2 = 402 + (109.12)2
x2 = 13,507.1744
32 + 42 − 52
Q. 10. cos A = ____________
2(3)(4)
−1
A = cos (0)
A = 90°
Q. 11. (i) |AC|2 = 42 + 62 − 2(4)(6) cos 60°
x = 116.22 cm
Max length = 116.22 cm
12
Active
ti Maths
th 3 Book
k 2 – St
Strands
d 1–5 – Ch 12
2S
Solutions
l ti
= 28
___
|AC| = √ 28 = 5.29
(ii)
A
53.8°
30°
(iii)
B
7
5
96.52°
5.29
9
C
52
+ 72 − 92
cos x = ____________
2(5)(7)
−1
x = cos (−0.1)
|BC|
|BC|
______
5.29
= ________
sin 30° sin 53° 8ʹ
5.29 sin 30°
|BC| = ___________
sin 53° 8ʹ
|BC| = 3.31
1
(iii) ABC = __ (5.29)(3.31) sin 96° 52ʹ
2
= 8.6922
1
ADC = __ (4)(6) sin 60°
2
= 10.3923
(iv)
Q. 14.
(iii)
Q. 13.
(i)
1
__
5 cm
(iii)
3 cm
5
3
7
52
+ 72 − 32
cos x = ____________
2(5)(7)
−1
x = cos (0.9286)
x = 22°
1
(iv) __ (5)(7) sin 22°
2
6.5556 square units
Q. 15.
(6)(8) sin 117° = 21.3842
2
1
__
(7)(12)sin 94° = 41.8977
2
63.2819 square
units
(i) 60 × 0.25 = 15 km
(ii) 50 × 0.25 = 12.5 km
(iii) 150°
(iv) x2 = 152 + 12.52 − 2(15)(12.5)
cos 150°
= 706.0095
7 cm
5 cm
(5)(7) sin 96°
2
17.4041 units square
(ii) 22°
|EG|
7
(i) ______ = ______
sin 56° sin 30°
7 sin 56°
|EG| = ________
sin 30°
= 12 cm
82 + 62 − 122
(ii) cos|∠EHG| = _____________
2(8)(6)
−1
|∠EHG| = cos (−0.4583)
|∠EHG| = 117°
x = 96°
7 cm
Area = 19.0845 units squared
Q. 12.
(i)
1
__
x = 26.57 km
9 cm
(ii) 96°
Q. 16.
(i) x2 = 1762 + 1352 − 2(176)(135) cos 38.71°
= 12161.66354
x = 110.28 km
≅ 110 km
(ii) Journey = 176 + 110 + 135 = 421 km
D 421
T = __ = ____ ≅ 1.5 hrs.
280
S
Total trip 1.5 + 1 + 2 = 4.5 hrs.
Active Maths 3 Book 2 – Strands 1–5 – Ch 12 Solutions
13
(iii) We need both journey times.
127
Car: ________ = 2.54 hrs = 2 hrs 32 mins
50 km/h
110
Limerick to Waterford
Helicopter: _________ = 0.39 hrs
280 km/h
176
____
= 0.63 hrs
Dublin to Limerick
280
Stopover
1 hr
Total time to Waterford including stopover = 2.02 hrs
= 2 hrs 1 min
The car arrives in Waterford 31 mins after the helicopter.
The helicopter must be in the air in 2 hrs from time of landing.
Max. length for meeting is 1hr 29 mins.
Q. 17.
33.072 + 33.072 − 35.422
(i) cos B = _______________________
2(33.07)(33.07)
−1
B = cos (0.4264)
B = 64.76°
1
(ii) __ (33.07)(33.07) sin 64.76°
2
494.61 m2
(iii) 494.61 × 4 = 1978.44 m2
(iv) 35.42 × 35.42 = 1254.5764 m2
Q. 18.
(i) |AC|2 = 352 + 402 − 2(35)(40) cos 120°
= 4225
|AC| = 65 m
(ii) 35 + 40 + 65 = 140 m
(iii) Use are of Δ.
1
1
__
(35)(40) sin 120° = __ (35)(⊥height)
2
2
40 sin 120° = 34.64 metres
or
sin 60° =
C
60°
40 m
B
x
___
40
x = 40 sin 60°
x = 34.64 metres.
Exercise 12.11
Q. 1.
14
50
(i) ____ p(10)2
360
= 13.89p cm2
150
(ii) ____ p(12)2
360
= 60p cm2
210
(iii) ____ p (18)2
360
= 189p cm2
Q. 2.
( )
40
(i) 2p(20) ____
360
= 4.44p cm
100
(ii) 2p(25) ____
360
= 13.89p cm
170
(iii) 2p(30) ____
360
= 28.33p cm
Active Maths 3 Book 2 – Strands 1–5 – Ch 12 Solutions
( )
( )
Q. 3.
Q. 4.
q
(i) ____ p (6)2 = 6p cm2
360
6q
____
=1
360
q = 60°
q
____
(ii)
p (12)2 = 68p
360
54
q = ____
____
360 144
q = 135°
q
(iii) ____ × 81p = 45p
360
45
q = ___
____
360 81
q = 200°
q
(iii) ____ × 2p (24) = 40p
360
40
q = ___
____
360 48
q = 300°
Q. 5.
1
(i) __(7)(7)sin 70° = 23.02 sq units
2
70
(ii) ____ × p(7)2 = 29.93 sq units
360
(iii) Shaded region 29.93 − 23.02
= 6.91 sq units
Q. 6. Δ
1
__
2
(13)(10) sin 40° = 41.7812
40
____
× p × 152 = 78.5398
360
Shaded = 78.5398 − 41.7817
Sector
q
(i) ____ × 2p (18) = 7p
360
7
q = ___
____
360 36
q = 70°
q
____
× 2p (6) = 4p
(ii)
360
4
q = ___
____
360 12
q = 120°
= 36.7586
≅ 36.8 sq units
Revision Exercises
Q. 1.
20
21
(a) sin a = ___
cots a = ___
29
29
20
21
cos b = ___
sin b = ___
29
29
sin 100°
sin a _______
_____
(b) (i)
=
7
15
7 sin 100°
sin a = _________
15
−1
a = sin (0.4596)
20
tan a = ___
21
21
tan b = ___
20
(iii) y2 = 12.162 + 202 − 2(12.16)(20) cos 120°
y2 = 791.0656
y = 28.13 cm
a = 27°
(ii)
Q. 2. (a)
15
= _______
sin 53° sin 100°
15 sin 53°
k = _________
sin 100°
k = 12.16 cm
k
______
sin a
cos a
p
−q
tan a
−p
___
q
__
1
(b) (i) __(x)(√3 ) sin 80° = 3
2
0.85290x = 3
x = 3.52
__2
__
(ii) y2 = 3.522 + √ 3 − 2(3.52)( √ 3 ) cos 80°
= 13.2730
y = 3.64
Active Maths 3 Book 2 – Strands 1–5 – Ch 12 Solutions
15
Q. 3. (a) a = 210 − 180 = 30°
1
sin a = __
2 __
√3
cos a = ___
2
1
___
tan a = __
√3
1
sin 210° = −__
2 __
√3
cos 210° = −___
2
1__
___
tan 210° =
√3 __
√3
m = −____
2
1
n = −__
2 __
√3 1
A = (−___,−__)
2
2
6
x
(b) (i) ______ = ______
sin 70° sin 65°
6 sin 70°
x = ________
sin 65°
x = 6.22 cm
1
(ii) __(6.22+5)(6)sin 45°
2
= 23.8012
(b)
5
(i) tan b = ___
12
12
(ii) cos b = ___
13
h2 = 52 + 122
h = 13
5
(iii) sin b = ___
13
(iv) sin 2b = 2 sin b cos b
( )( )
5 12
= 2 ___ ___
13 13
120
____
=
169
|OA| = 12 × 2
Q. 5.
= 24 km
|OB| = 16 × 2
= 32 km
|AB|2 = 242 + 322 −2(24)(32) cos 90°
|AB|2 = 1600
|AB| = 40 km
12
Q. 6. (a) |PQ| sin 30° = ___
x
x sin 30° = 12
12
x = ______
sin 30°
x = 24
= 24 sq units
Q. 4. (a)
(i) sin A = cos 60°
1
sin A = __
2
A = 30°
|PR|
(ii) tan B = 1
P
B = 45°
(iii) sin C = cos C
C = 45°
3
(iv) sin D cos 30° = __
4
__
3
√
3
sin D ___ = __
4
2
3
2__
sin D = __ × ___
4 √3
3__
= ____
2√
3
__
3
√
= ___
2
D = 60°
( )
16
Active Maths 3 Book 2 – Strands 1–5 – Ch 12 Solutions
R
17
24
50°
Q
|PR|2 = 242 + 172 − 24(17) cos 50°
= 602.7427
|PR| = 24.5508
≅ 25
Q. 8.
(b)
(i)
50°
15 cm
y
40°
x
x
___
= sin 50°
15
y
___
= sin 40°
15
x = 15 sin 50°
β
d
First, John should set up the
clinometer at a point, x, a distance
d from the base of the wall. The
value of d can be found using the
measuring tape. From x, John will
measure the angle of elevation, b,
of the billboard and the angle of
elevation, a, of the bottom of the
billboard.
y = 15 sin 40°
x = 11.4907
y = 9.6418
Area = 11.4907 × 9.6418
≅ 111 sq units
Q. 7.
α
x
(i) r = 2 cm
(ii)
60°
(ii) _____ × p × 22
360°
2.09 cm2
1
(iii) __(10)(10) sin 60°
2
43.30 cm2
h
__
= tan 23°
5
h = 5 tan 23°
h = 2 m = 200 cm
a+h
_____
= tan 45°
5
a + h = 5 tan 45°
(iv) 43.30 − 2.09(2)
a+h=5
= 39.12 cm2
a+2=5
a = 3 m = 300 cm
Q. 9.
(i) x2 = 62 + 102 − 2(6)(10) cos 47
x2 = 54.1602
x = 7.36 m
(ii)
(iii)
2.42
α
10
2.42
tan a = ____
10
a = tan−1
a = 13.6040
22°
6m
≅ 14°
x
tan 22 = __
6
x = 6 tan 22
x = 2.42 m
Active Maths 3 Book 2 – Strands 1–5 – Ch 12 Solutions
17
Q. 10.
(i)
C
(ii)
L
D
30 m
20°
15°
B
A
|AB|
(ii) cos 15° = _____
30
|AB| = 30 cos 15°
25
22
60°
X
sin(x)
_____
25
sin 60
= ______
22
25 sin 60
sin x = ________
22
|AB| = 28.98 m
|CB|
tan 35° = ______
28.98
|CB| = 28.98 tan 35°
x = (79.8)°
(iii) |∠ACF| = 180° − (60 + 79.8)
= (40.2)°
Hill + tower = 20.29 metres
(iv)
Hill =
D
30
20
B
A
x
sin 15 = ___
30
x = 7.76
D
Tower = 20.29 − 7.76
= 12.53 metres
(i)
40.2
?
15°
Q. 11.
C
C
18
60
79.8
x
x
________
sin (40.2)
E
18
= ______
sin 60
18 sin (40.2)
x = ___________
sin 60
= 13.4
25 cm
22 cm
A
F
C
18
20 cm
18 cm
D
E
Active Maths 3 Book 2 – Strands 1–5 – Ch 12 Solutions
|DE| = 13.4 cm