Chapter 21 Exercise 21.1
Q. 1.
________
(i) x = √122 + 52 = 13
Q. 7.
_________
(ii) x = √122 + 352 = 37
__________
(iii) x = √
1012
(iv) x = √
412
−
202
_________
−
402
_________
h
= 99
27°
=9
100 m
(v) x = √612 − 112 = 60
h
tan 27° = ____
100
_________
(vi) x = √652 − 632 = 16
Q. 2.
100 tan 27° = h
(i) x = 8 sin 31° = 4.1
5.5
(ii) x = _______ = 14.7
cos 68°
12
(iii) x = ______ = 15.9
sin 49°
4
(iv) x = _______ = 5.9
tan 34°
So, use the tangent function as shown
above to evaluate the height.
Q. 8.
33
(v) x = 8.5 cos 18° = 8.1
39°
(vi) x = 5 tan 53° = 6.6
Q. 3.
x = 33 sin 39°
()
5
(i) A = cos−1 __ = 56°
9
10
___
(ii) A = sin−1
= 56°
12
4
(iii) A = sin−1 __ = 30°
8
15
(iv) A = tan−1 ___ = 56°
10
4
(v) A = tan−1 __ = 34°
6
2
(vi) A = cos−1 __ = 71°
6
∴ h = 33 sin 39° + 1.5
( )
()
( )
()
()
Q. 4.
y=
Q. 9.
32°
y
14
(i) sin 32° = ___
x
14
x = ______
sin 32°
4
_______
x ≈ 26.42 metres
y ≈ 22 metres
4
___
Q. 10.
10
Q. 6. cos 30° = _____
|AC|
10
|AC| = _______
cos 30°
|AE| =
x
14
( 12 ) = 18°
tan−1
|AD| =
32°
(ii) y = 26.42 cos 32°
(i) y = 4
(ii) A =
= 2,227 cm
tan 27°
y = 7.9
x = 4.0
Q. 5.
= 22.27 metres
4
tan 27° = __
y
x
tan 53° = __
3
x = 3 tan 53°
x
(
(i) No. For a given opposite length,
doubling the adjacent length does
not halve the angle. It halves the
tangent of the angle.
(ii)
)
10
______
cos 30°
_______
cos 30°
(( ))
10
______
cos 30°
_______
cos 30°
________
cos 30°
x
__
=
80√3
_____
9
30°
12
Active Maths 2 (Strands 1–5): Ch 21 Solutions
1
x
tan 30° = ___
12
x = 12 tan 30°
Q. 13.
(i)
__
x = 4√ 3
__
∴ height = 4√ 3 + 1.4
30
= 8.33 metres
Q. 11.
(i)
68°
h
x
27°
tan 27° =
30
tan 68° = ___
x
81
30
x = _______
tan 68°
h
___
87
h = 81 tan 27°
x = 12.12
h = 41 metres
(ii)
30
33°
y
30
tan 33° = ___
y
103
30
y = _______
tan 33°
q
y = 46.20
81
∴ |BC| = 46.2 − 12.12
( 81 )
103
q = tan−1 ____
q = 52°
Q. 12.
2
(i) sin 30° = __
x
x=
2
______
= 34.08 metres
Q. 14.
sin 30°
x = 4 metres
34.08
(ii) ______ = 3.4 m/s
10
8 4
(i) tan A = __ = __
6 3
Draw a right triangle with
opposite length 4 units and
adjacent length 3 units.
(ii) y = 2(4 cos 30°)
= 8 cos 30°
= 6.93 metres
4 units
A
3 units
Use a ruler, set square and
compass.
(Diagram shown is not to scale.)
2
Active Maths 2 (Strands 1–5): Ch 21 Solutions
(ii) By Pythagoras, hypotenuse length
is 5 units.
4
∴ sin A = __
5
3
__
cos A =
5
Q. 15.
(iv) h = h
x tan 42° = (7 − x) tan 48°
x tan 42° = 7 tan 48° − x tan 48°
x tan 42° + x tan 48° = 7 tan 48°
x (tan 42° + tan 48°) = 7 tan 48°
(i) Similar method as to Q.14.
7 tan 48°
x = _______________
tan 42° + tan 48°
(v) x = 3.865….
∴ h = (3.865….)x tan 42°
3 units
h = 3.48 m
h = 348 cm
A
Q. 17.
4 units
(i)
(ii) Hypotenuse length is 5 units
3
sin A = __
5
4
cos A = __
5
Q. 16.
10
h
q
(i) q = 90° − 42° = 48°
h = 10 sin q
(ii) 7 – x metres
(ii)
(iii)
h
10
y
q
10
h
q
h
10
10
42°
M
y
x
N
h
__
tan 42° = x
h = x tan 42°
y = 10 cos q
∴ l = 10 cos q + 10 + 10 cos q
l = 10 + 20 cos q
1
(iii) Area = __(10 cos q)(10 sin q)
2
+ 10(10 sin q)
1
+ __(10 cos q)(10 sin q)
2
= 50 cos q sin q + 100 sin q
h
+ 50 cos q sin q
48°
7–x
h
tan 48° = _____
7− x
h = (7 − x) tan 48°
∴ A = 100 sin q + 100 sin q cos q
(iv) A = 100 sin 60° + 100 sin 60°
cos 60°
__
= 75√ 3 cm2
(v) A = 100 sin 45° + 100 sin 45°
cos 45°
__
= 50 + 50√ 2 cm2
Active Maths 2 (Strands 1–5): Ch 21 Solutions
3
Q. 18.
(i) SU is a tangent to the circle and
[OU] in the corresponding radius.
Q. 3.
A tangent and corresponding
radius make a 90° angle with
each other.
(ii) 90° − q
(iii) 180° in a triangle. So |∠OSU|
Q. 4.
(ii) 16°52ʹ
(vii) 7.38ʹ
(iii) 68°10ʹ
(viii) 13°40ʹ
(iv) 132°17ʹ
(ix) 7.30ʹʹ
(v) 36°15ʹ
(x) 56.27ʹ
(i) A = 90° – 44°44ʹ = 45°16ʹ
= 59° 12ʹ
=q
(iii) C = 72°14ʹ + 92°51ʹ
(iv) |UC| = sin q
= 165° 5ʹ
1
(iv) D = __(180° − 30°40ʹ)
2
(v) |OC| = cos q
(vi) Pythagoras:
= 74°40ʹ
sin2 q + cos2 q = 1
Q. 5.
O
q
x
________
(i) |AC| = √ 92 + 122 = 15 km
15
(ii) tan 36.87° = _____
|DC|
15
|DC| = _________
tan 36.87°
1
U
(vi) 6°38ʹ
(ii) B = 180° − 50°48ʹ − 70°
= 180° − 90° − (90° − q)
(vii)
(i) 6°45ʹ
|DC| ≈ 20 km
S
_________
1
tan q = __
x
(iii) |AD| = √ 152 + 202 = 25 km
1
x = _____
tan q
∴ Length = 9 + 12 + 20 + 25
= 66 km
1
∴ |SU| = _____
tan q
Q. 6.
(i) As |∠CBA| = 90°
⇒ a is acute (only 180° in a triangle)
Exercise 21.2
Q. 1.
Q. 2.
(ii) If 45° < a < 90°
(i) 2°30ʹ
(vii) 24°45ʹ
(ii) 2°15ʹ
(viii) 36°15ʹ
(iii) 2°45ʹ
(ix) 55°08ʹ
(iv) 25°24ʹ
(x) 0°20ʹ
(v) 4°12ʹ
(xi) 0°45ʹ
(vi) 15°30ʹ
(xii) 0°24ʹ
(i) 2.5°
(vi) 33.55°
(ii) 10.67°
(vii) 42.17°
(iii) 25.83°
(viii) 58.75°
(iv) 70.37°
(ix) 56.5°
(v) 11.6°
(x) 82.2°
For such values of a, the opposite
length exceeds the adjacent
length (i.e. |AB| > |BC|).
(iii) sin a: opposite length is always
less than hypotenuse length.
cos a: adjacent length is always
less than hypotenuse
length.
(iv) Choose say a = 30°
sin 2a = sin 60° ≈ 0.866
2 sin a = 2 sin 30° = 1
0.866 ≠ 1
So statement (by this example) is
false.
4
Active Maths 2 (Strands 1–5): Ch 21 Solutions
(i)
Sun
r = sin 60°16ʹ 34ʹʹ
Moon
q
+ 960 sin 60°16ʹ 34ʹʹ = r
3.84 × 108
Q. 7.
x
r = (sin 60°16ʹ 34ʹʹ − 1)
= −960 sin60°16ʹ 34ʹʹ
80° 50’ 58” Earth
−960 sin 60° 16ʹ 34ʹʹ
r = ___________________
sin 60° 16ʹ 34ʹʹ – 1
q = 90° − 89° 50ʹ 58ʹʹ
r = 6,336.211…
= 9ʹ 2ʹʹ
r ≈ 6,336 km
3.84 × 108
(ii) sin 9ʹ2ʹʹ = __________
x
(iv) 7,296 km
3.84 × 108
x = __________
sin 9ʹ 2ʺ
(v) 2p(7,296)
x = 2,445,718,857
≈ 45,842 km
45,842
(vi) _______ = 1.697… hours
27,000
= 2.445718857 × 109 in metres
Distance
(iii) Time = ________
Speed
2.445718857 × 109
= __________________
3.0 × 108
= 8.1523….
≈ 1 hr 42 mins
Exercise 21.3
≈ 8.15 seconds
Q. 1.
(iv) r = 2.445718857 × 109
Circumference = 2pr
(i) Construct equilateral triangle of
side length 2 units.
Divide in two from top vertex.
≈ 1.54 × 1010 metres
(v) 365_14 days = 365.25 × 24 × 60
× 60
30°
= 31,557,600 seconds
1010
1.54 ×
(vi) ___________
31,557,600
2
x
≈ 488 m/s
Q. 8.
(i) 90° − 60°16ʹ 34ʹʹ = 29°43ʹ 26ʹʹ
60°
(ii) r + 960 km
(iii)
1
A
6’
60° 1
34”
B
__
x = √3
__
r + 960
O
r
sin 60°16ʹ 34ʹʹ = _______
r + 960
√3
___
1
sin 30° = __
2__
3
√
___
cos 30° =
2
__
1__
tan 60° = √ 3 tan 30° = ___
√3
Construct a right-angled isosceles
triangle with equal sides length
one unit.
So: sin 60° =
r
_______
By Pythagoras: x = √22 − 12
2
1
__
cos 60° =
2
Active Maths 2 (Strands 1–5): Ch 21 Solutions
5
z
tan 60° = __
9
9 tan 60° = z
45°
__
9√ 3 = z
1
x
Q. 2.
(i) 60°. ΔCAB in equilateral.
(ii) RHS
45°
__
(iii) √ 3 . By Pythagoras.
1
_______
By Pythagoras: x = √
12
__
+
x = √2
12
1__
So: sin 45° = cos 45° = ___
√2
tan 45° = 1
A
sin A
30°
45°
60°
1
__
1__
___
√3
___
2
cos A
2
tan A
1__
___
1
√3
√3
2
__
(iii) 4 + h = √ 3 h
__
4 = h(√ 3 − 1)
2
_______
__ 4
__
√3 − 1
=h
__
4(√ 3 + 1)
_________
3−1
=h
__
2(√ 3 + 1) = h
Q. 4.
x
tan 60° = __y
6
tan 60° = __
x
y = 102 − 52
y2 = 75
y = √ 75
6
x = _______
tan 60°
y = 5√ 3
∴ x = 2√ 3
___
__
__
__
5
__
cos 30° = z
__
√3 = __
5
___
z
2
10
__
z = ___
√3 __
10√ 3
z = _____
3
y
__
(iii) sin 60° = ____
6√3
__
y
√3 = ____
___
__
2
6√3
∴y=9
y=
2√3
_______
tan 60°
∴y=2
Q. 5.
__
x = 6√ 3
cos 60°
__
6√ 3
x = ____
2__
x = 3√ 3
6
(1)(√ 3 ) =
(i) h = |DC|
x
(ii) sin 30° = ___
10
1 ___
x
__
=
2 10
∴x=5
2
√3
___
h
(ii) tan 30° = ______
4+h
h
1__ ______
___
=
4
+
h
3
√
2
1
__
__
Q. 3.
__
√2
1__
___
√2
√3
___
(iv)
__
__
1
__
Active Maths 2 (Strands 1–5): Ch 21 Solutions
(i)
x
45°
x
45°
__
(ii) x, x and √ 2 x
(iii)
(x) By considering how the ratio
adjacent: hypotenuse changes as
a increases.
60°
8√6 – x
Let a = ∠DAB, then ∠EAB, then
∠FAB etc….
30°
x
__
The adjacent length |AB|
remains the same, while the
hypotenuse lengths |AD|, |AE|,
|AF|, etc increase. Hence, cos a
decreases as a increases.
__
(iv) x, 8√ 6 − x, 16√6 − 2x
(v) 1 − 3 = −2
__
8√6 − x
________
(vi) tan 30° =
x
8√6 − x
1__ ________
___
=
x
√3
__
__
Q. 7.
__
Q. 8.
__
x (1+√3 ) = 24√2
__
24√ 2__
x = _______
1 + √3
x ≈ 12.4
Q. 9.
(vii) 12.4 − 1.7
Q. 10.
= 10.7 metres
Q. 6.
(ii)
1__ 2 __
1 1
+ ___
= + __ = 1
2 2
2
2
3
___
2
1__
___
3
__
3 1
12
+ __ + ( 3 )2 = __ + __ + 3 = 4
4 4
2
__
2
2
2
3
1
1 1 3
+ __ + ___ = __ + __ + __
2
2
3 4 4
4
= __
3
2
_
1
1__
2
___
LHS = __ = ___
= RHS
√3
___
√3
2
(i) Diagram 2. She’s increasing the
opposite length as the adjacent
length remains constant.
_______
1__ 2
___
__
x = 24√2 − √ 3 x
__
(√ ) (√ )
(√ ) ( ) √
(√ ) ( ) (√ )
Q. 11.
(i) O
4
__
√42 + 82 = 4√5
30°
(iii) q = 90° − a and b = 90° − a
M
(iv) They both contain a 90° angle
and q = b.
|BC|
12
__
(v) _____ = ___
1
√3
4
tan 30° = __
x
4
x = _______
tan 30°
∴ |BC| = 4√ 3
(ii)
__
(vi) |AB| = √(4 √ 3 )2 + 122 = 8√ 3
1
(vii) Yes. In Δ ADE, sin a = __
2
__
3 __
4√__
1
=
In ΔABC, sin a = ____
2
8√ 3
(viii) |AB| = 8 units
|BC| = 10 units
_______
___
(ix) |AD| = √ 82 + 22 = 2√17
_______
__
|AE| = √ 82 + 42 = 4√ 5
_______
|AF| = √ 82 + 62 = 10
_______
__
|AG| = √ 82 + 82 = 8√2
N
__
|MN| = 4√ 3
__
____________
__
x
___________
__
___
√42 + (4√3 )2 = √64 = 8
(iii) 4 + 8 = 12
__
1 __
(iv) __(8√3 ) (12) = 48√3 units2
2
Revision Exercises
Q. 1.
_______
(i) x = √ 62 − 42 = 4.5
_________
(ii) x = √ 42 − 2.62 = 3.0
4
(iii) tan 26.6° = __
x
⇒
4
x = ________ = 8.0
tan 26.6°
Active Maths 2 (Strands 1–5): Ch 21 Solutions
7
(iv) x = 4.3 sin 56.4° = 3.6
(ii)
B
6.8
(v) cos 47.3° = ___
x
6.8
________
45°
= 10.0
⇒
x=
⇒
x = 3.6 tan 33.4° = 2.4
cos 47.3°
x
(vi) tan 33.4° = ___
3.6
3
45°
A
(vii) x = 6 sin 45° = 4.2
3
_______
H
|AB| = √ 32 + 32 = 4.24 m
_______
(viii) x = √ 32 + 42 = 5
(iii)
B
()
( 43 ) = 41°
( 46 ) = 42°
( 35 ) = 59°
( 43 ) = 37°
2
(i) tan−1 __ = 27°
4
Q. 2.
(ii) cos−1 __
(iii) sin−1 __
3
(iv) tan−1 __
(v) tan−1 __
Q. 3.
Q
40°
_______
y
36°
x
(iv) 2(4.24) + 3 + 2(3) + 2(3.16)
+ 3.16
= 26.96 m
R
4.1
tan 40° = ___
y
4.1
y = _______
tan 40°
(v) 36 × 3.4 × 26.96 = €3,299.904
10
___
y ≈ 4.886
9
4.886
tan 36° = ______
x
x=
H
|BD| = √ 12 + 32 = 3.16 m
4.1 T
P
1
D
Q. 5.
(3,299.904) = €3,666.56
(i) B = tan−1 (0.2908)
2B = 2 tan−1 (0.2908)
4.886
_______
tan 36°
tan 2B = tan [2 tan−1 (0.2908)]
x ≈ 6.725
= 0.635
1
∴ Area = __(4.1 + 6.725)(4.886)
2
(ii)
B
≈ 26.4 units2
h
Q. 4.
(i) |DH| = |HE| as Δ BDH
≡ ΔBHE
(RHS)
⇒ |AH| = |HC| as |AD| = |EC|
⇒ ΔABH ≡ ΔCBH (SAS)
23°35’
A
97
h
tan 23°35’ = ___
97
h = 97 tan 23°35’
h ≈ 42.34 m
∴ |BT| = 42.34 m
8
Active Maths 2 (Strands 1–5): Ch 21 Solutions
T
(iii)
|AC|2 = 142 + 202
B
|AC|2 = 596
42.34
____
|AC| = 2 √ 149
____
y
∴ r = √ 149 cm
Q. 10.
80° 30’
h
(i) tan 51° = __
x
x tan 51° = h
T
C
h
tan 30° = _________
14.75 + x
42.34
sin 80° 30ʹ = ______
y
42.34
y = __________
sin 80° 30ʹ
(14.75 + x) tan 30° = h
(ii) h = h
∴ |BC| = 42.93 m
Q. 6.
Q. 7.
__
(
x tan 51° = 14.75 tan 30°
)
10√ 3
tan−1 _____ = 60°
10
+ x tan 30°
x (tan 51° − tan 30°)
= 14.75 tan 30°
14.75 tan 30°
∴ x = _______________
tan 51° − tan 30°
9
(iii) x = 12.95 m
30°
x
(iv) h = 12.95 tan 51°
9
tan 30° = __
x
x=
h = 15.99…
9
_______
h ≈ 16 m
tan 30°
__
x = 9√3
__
(v) 2pr = 25.13
__
r = 3.999…
Perimeter = 6(9√ 3 ) = 54√ 3 cm
r≈4m
+
8
___
√152 – 62 = 3√21
√152 – 62 = 3√21
Q. 8.
15
___
1
Area = 8(3√ 21 ) + __ (6) (3√21 )
2___
___
= 24√ 21 + 9√ 21
___
= 33 √ 21
≈ 151.22 cm2
Q. 9.
1
__
|AB||BC| = 140
2
|AB||BC| = 280
6
(vi) The distance x can now be
calculated once the distance
(x − 4) is measured directly.
One angle of elevation
measurement will then be
sufficient to calculate the height,
by use of the tangent function.
(vii) (x − 4) in measured as 8.95.
h
tan 51° = __
x
h
______
tan 51° =
12.95
h = 12.95 tan 51°
∴ h = 15.99 ≈ 16 m
|AB| (14) = 280
|AB| = 20 cm
Active Maths 2 (Strands 1–5): Ch 21 Solutions
9
Q. 11.
________
|AB|
(i) tan 61.8° = _____
30
30 tan 61.8° = |AB|
|AB| = 55.94…
|AB| ≈ 56 m
(ii) sin |∠BDA| =
= 36.886…
56
____
≈ 37 cm
120
(iv) |BD| = 37 − 3 = 34 cm
|∠BDA| = 27.818…
|∠BDA| ≈ 28°
Q. 12.
___
(iii) |AC| = √ 102 − 32 = √91 cm
|AB|
___
tan (58°3’ + 17°) = _____
91
√
___
|AB| = √ 91 tan (75° 30’)
(i) |∠DAC| = 60°
Q. 16.
h
(i) tan 51°36’ = ___
15
(ii) |BD| = 2 m
(iii)
h = 15 tan 51°36ʹ
C
h = 18.925…
h ≈ 18.93 m
18.93
(ii) tan 28°30ʹ = ______
15 + x
18.93
15 + x = __________
tan 28° 30ʹ
4
x ≈ 19.86 m
60°
Q. 13.
B
A
h
4
sin 60° = _____
|BC|
4
|BC| = ______
sin 60°
4
A
1
|BC| ≈ 4.62 m
_______
h = √42 + 12
(iv)
___
h = √17
1
___
∴ cos A = ____
√17
Q. 14.
tan 30° 15’ =
(i) Incorrect. tan 50° > 1
Q. 17.
20
_______
20 + d
20
20 + d = __________
tan 30° 15ʹ
(ii) Correct. sin 60° ≠ 2 sin 30°
(iii) Incorrect. cos 21° < cos 20°
(iv) Correct
d = 14.29…
d ≈ 14 m
Q. 15.
(i) |∠ACD| = 17°
|AD|
(ii) sin 17° = _____
10
10 sin 17° = |AD|
|AD| = 3 cm
10
Active Maths 2 (Strands 1–5): Ch 21 Solutions
__________
√4.622 − 42 ≈ 2.31 m
60°
2 cm
2 cm ⇒
60°
60°
2 cm
∴ sin 60° =
__
√3
___
2
√3
30°
2 cm
60°
1 cm
Q. 18.
Leaning Tower of Pisa
2pr = 7.07
r ≈ 1.125
x
tan 60° = _______
71.125
71.125 tan 60° = x
55.863
x = 123.192…
∴ h = 123.192… + 1.72
θ
∴ h ≈ 125 m
3.9
(
55.863
q = tan−1 _______
3.9
)
Q. 20.
(a)
l
30
q ≈ 86.006°
105
__________
Suurhusen church tower
l = √ 302 + 1052
l ≈ 109.2 cm
h
(b) (i) tan 60° = __
x
27.37
h = x tan 60°
__
h = √3 x
∝
h
(ii) tan 30° = ______
24 − x
2.47
(
27.37
a = tan−1 ______
2.47
)
(24 − x) tan 30° = h
24 −
______
__ x
√3
a ≈ 84.843°
h=h
As a < q
__
24 − x
__
√3 x = ______
√3
⇒ The Suurhusen church tower is
more tilted.
Q. 19.
=h
3x = 24 − x
(a) Ultan’s measurement will be
used to calculate the radius of
the Spire, which will allow for
the height of the Spire to be
calculated via use of the tangent
function.
4x = 24
x=6
__
h = 6√ 3 metres
Q. 21.
(a)
(b)
x
x
h
h
73
60°
1.72
.68
°
1.62
56.38
71.125
Active Maths 2 (Strands 1–5): Ch 21 Solutions
11
x
(b) tan 73.68° = ______
56.38
x = 56.38 tan 73.68°
h = 56.38 tan 73.68° + 1.62
h ≈ 194.1750 metres
(c) Second: 194.0727 metres
Third: 194.1149 metres
194.1750 + 194.0727 + 194.1149
(d) _______________________________
3
= 194.1208…
≈ 19,412 cm
12
Active Maths 2 (Strands 1–5): Ch 21 Solutions