Quadratic Equations & Functions Exercise 1A
4 (f) x2 7 6x +
49 2 = 0
x2 7 6x =
49 2
(x 7 3)2 (
7 3)2 =
49 2
(x 7 3)2 =
49 9
49 2
x 7 3 = ±
7 1
x = 0.81 or 0.05 (g) x2 + 0.6x 1 = 0
x2 + 0.6x = 1
(x + 0.3)2 (0.3)2 = 1 (x + 0.3)2 = 0.09 + 1 x + 0.3 = ± 1.09
x = 0.74 or 1.34 (h) x2 4.8x + 2 = 0
x2 4.8x = 2
(x 2.4)2 (2.4)2 = 2 (x 2.4)2 = 5.76 2 x 2.4 = ± 3.76
x = 4.34 or 0.46 5 (c) (x + 2)(x 5) = 4x
x2 3x 10 4x = 0 x2 7x 10 = 0 (x
2 7)2 (
2
7)2 = 10
(x 2 7)2 =
4
49 + 10
x 2 7 = ±
4 89
x = 8.22 or 1.22 (d) x(x 4) = 2(x + 7)
x2 4x 2x 14 = 0 x2 6x 14 = 0 (x 3)2 (3)2 = 14 (x 3)2 = 9 + 14 x 3 = ± 23
Exercise 1B
3 (e) (2x + 3)(x 1) x(x + 2) = 0 2x2 + x 3 (x2 + 2x) = 0 x2 x 3 = 0
Using a = 1, b = 1 and c = 3, x = 2(1)
) 3 )(
1 ( 4 ) 1 ( ) 1
( 2
= 2.30 or 1.30
(f) (4x 3)2 + (4x + 3)2 = 25
16x2 24x + 9 + (16x2 + 24x + 9) = 25 32x2 = 7
x2 = 32
7 x = ±0.468
4 (b)
4
3x2 + 2x 2 1 = 0
Using a = 4
3, b = 2 and c = 2 1,
x =
4) (3 2
2) )( 1 4 (3 4 2
2 2
= 0.230 or 2.90 (c) 5x 7 = x2
x2 5x + 7 = 0
Using a = 1, b = 5 and c = 7, x = 2(1)
) 7 )(
1 ( 4 ) 5 ( ) 5
( 2
=
2 3 4
There is no solution since is not defined . 3 Exercise 1D
4 2
) 1 (
) 3 (
x
x
x = 5
3 5(x2 3x) = 3(x2 + 2x + 1)
2x2 21x 3 = 0
x =
) 2 ( 2
) 3 )(
2 ( 4 ) 21 ( ) 21
( 2
= 10.6 or 0.141
5 (f)
1 7
x
3 1
x x =
2
1
) 3 )(
1 (
) 1 )(
1 ( ) 3 ( 7
x x
x x
x =
2 1
2(7x + 21 x2 + 1) = x2 + 2x 3
3x2 12x 47 = 0
x =
) 3 ( 2
) 47 )(
3 ( 4 ) 12 ( ) 12
( 2
= 6.43 or 2.43
5 (g)
2 5
x = 2 2
) 2 (
4
x
2 5
x + 2
) 2 (
4
x = 2
2
) 2 (
4 ) 2 ( 5
x
x = 2
5x 6 = 2(x2 4x + 4)
2x2 13x + 14 = 0
x =
) 2 ( 2
) 14 )(
2 ( 4 ) 13 ( ) 13
( 2
= 5.14 or 1.36
5 (h)
1 5
x + 2
) 1 ( x
x = 1 2
) 1 (
) 1 ( 5
x
x
x = 1
6x 5 = (x2 2x + 1)
x2 8x + 6 = 0
x =
) 1 ( 2
) 6 )(
1 ( 4 ) 8 ( ) 8
( 2
= 7.16 or 0.838
8 (i) x
60 pages/min (ii)
2 60
x pages/min (iii)
x 60 +
2 60
x = 144 )
2 (
60 ) 2 ( 60
x x
x
x = 144
120x + 120 = 144(x2 + 2x)
24: 5x + 5 = 6x2 + 12x
6x2 + 7x 5 = 0 (Shown) (iv) (2x 1)(3x + 5) = 0
x = 2
1 or x = 1 3 2
(v) Reject x = 1 3
2 as x > 0
Required time taken = (2 2
1 60) 144 = 6 mins
11 (i) Speed of coach = x 30, so time taken = 30 700
x h
Time taken by car = x 700 h Total time taken = 20 h
30 700
x +
x
700 = 20
20:
) 30 (
) 30 ( 35 35
x x
x
x = 1
70x 1050 = x2 30x
x2 100x + 1050 = 0 (Shown)
x = 2(1)
) 1050 )(
1 ( 4 ) 100 ( ) 100
( 2
= 88.079 or 11.921 = 88.08 or 11.92 (iii) Reject x = 11.92 as x = 30 becomes < 0
Time taken for return journey =
079 . 88
700 = 7.95 h
Exercise 1E
1 (e) y = (3 x)(x + 2) (f) y = (2 x)(4 x)
2 (d) y = (x + 1)2 3 (f) y = (x 4)2 1
Line of symmetry is x = 1. Line of symmetry is x = 4.
7 (i) x2 8x + 5 = (x 4)2 (4)2 + 5 (ii)
= (x 4)2 11 (iii) Min. point = (4, 11).
(iv) Line of symmetry is x = 4.
Exercise 2A
6 (g)
5
1(3g + 4) 3
1(g + 1) 1 3
1(g + 5)
15: 3(3g + 4) 5(g + 1) 15 5(g + 5) 4g + 7 5g 10
9g 17 g 1
9 8
2 y
x y = (3 x)(x + 2)
0 6
3
y
x y = (2 x)(4 x)
0 4
8
2
0.732
2.73
y
x y = (x + 1)2 3
0
3 2 (1, 3)
1
(4, 1)
17 y
x
y = (x 4)2 1
0 4
4 5
7.32 0.683
y
x y = (x 4)2 11
0
(4, 11)
11
6 (h) 4(
3 h +
4 3) < 3(
2 h 5)
4( 12 9 4 h
) < 3(
2
10 h )
6: 2(4h + 9) < 9(h 10) 8h + 18 < 9h 90
h < 108 h > 108
7 6
1(2 p) 3 10
p
30: 5(2 p) 90 3p
5p 80 3p 8p 80 p 10
Largest possible value of p = 10 Exercise 2B
5 (c) 3x 3 < x 9 < 2x
3x 3 < x 9 and x 9 < 2x
2x < 6 x < 9
x < 3 x > 9
9 < x < 3 (d) 2x x + 6 < 3x + 5
2x x + 6 and x + 6 < 3x + 5
x 6 2x < 1
x >
2 1
2
1 < x 6
8 Let x be the no. of correct answers, then (30 x) are the no. of wrong answers.
Given: 5x + (2)(30 x) 66 5x 60 + 2x 66 7x 126
x 18
Max. correct answers obtained = 18
10 2
1x 4 >
3
1x and 6
1x + 1 <
8 1x + 3 3x 24 > 2x 4x + 24 < 3x + 72 x > 24 x < 48
24 < x < 48
So, x = 29, 31, 37, 41, 43, 47
0.5 6
9 3
13 (d) 3d 5 < d + 1 2d + 1
3d 5 < d + 1 and d + 1 2d + 1
2d < 6 d 0
d < 3 d ≥ 0
0 d < 3
14 (d) d 5 <
5 2d
2 d +
5 1
d 5 <
5
2d and
5 2d
2 d +
5 1
5
3d < 5
10 d
5 1
d < 8 3
1 d ≥ 2
2 d < 8 3 1
15 (c) 4 7 3x 2
11 3x 5
3
11 ≥ x ≥ 3 5
1
3
2 x 3 3 2
Required integers are 2 and 3.
(d) 10 < 7 2x 1
17 < 2x 8
2
17 > x ≥ 4
4 x < 8 2 1
Required integers are 4, 5, 6, 7 and 8.
Exercise 3A
6 (e) (e3)5 (e2)4 = e15 e8 = e7
(f) (4f 6)3 (2f 3)3 = 64f 18 (8f 9) = 8f 9 7 (c) (8e5f 3)2 (e3f)3 = 64e10f 6 (e9f 3) = 64ef 3
(d) 16g8h7 (2g3h2)3 = 16g8h7 (8g9h6) = 2g1h = g
h 2
8 (c)
4
2
3 3
f
e
11
27 9
f
e = 8
81 12
f
e 9
11
27e
f = 3e3f 3
(d)
6 3 2
h g
3 2
5
2 3
h
g = 18
12
h
g ( 15
6
27 8 g
h ) = 3 12 27
8 h g
Exercise 3B
6 (f) (1000)3
2
= (3 1000)2 = (10)2 = 100
7 (f) 5
3 )
( 1
f =
3 5
1 f
= f 3
5
8 (d) 10d = 0.01
10d = 102 d = 2
10 (g) 1 2
2 2 3 5
) (
) ( ) (
n m
m n m
= 2 2
4 3
5 )
(
n m
m n m
= m5 + (4) (2)n3 2 = m3n (h) (5p)3 10p 7p2 + 63
p = 125p3 70p3 + 6p3
= 61p3
11 (e) (e3f 4) 2
1
= e2
3
f 2 = 2
2 3
f e
(f) (g3
2
h 5
4
)2
3
= gh 5
6
=
5 6
h g
12 (e) (4j4k)2
1
2h3k 2
1
= 2j2k2
1
2h3k 2
1
= 3
2
h k j
(f) (m3n 4
1
)4 5 32m4n8 = m12n1 2m5
4
n 5
8
= 2
5 3 5 56
n m
14 Using A = $5800, P = $5000 and n = 5 years.
5800 = 5000(1 + 100
r )5
(1 + 100
r )5 = 5000 5800
1 + 100
r = 5
1
50)
(58 r = 3.01%
15 Let P be the original sum of money, then r = 4
2 .
4 = 1.05 and n = 4.
P + 96.60 = P(1 + 100
05 . 1 )4
P(1 + 100
05 .
1 )4 P = 96.60
P[(1 + 100
05 .
1 )4 1] = 96.60 P = $2264.09
Exercise 3C
5 Required % = 6
12
10 1500
10 91 . 5
100% = 0.394 106 %
= 3.94 105 %
6 (g) 3
5 4
10 3
10 16 . 2 10 37 . 4
= 15.29 102
= 1.53 101
(h) 6 8
10
10 5 . 3 10 2 . 7
10 4 . 2
= 0.0033496 102
= 3.35 105 7 (d) 3 9.261106 = 2.1 102
(f) 2 3
3 2
10 4 . 3 10 2
10 5 . 2 10 8
= 1.99 105
9 Given: x = 2 103, y = 7 104
x + 8y = 2 103 + 8(7 104) = 7.6 103 12 (i) Speed = 300 000 000 m/s = 3 108 m/s
(ii) Time taken =
m/s 10 3
km 10 5 . 778
8 6
=
m/s 10 3
m 10 10 5 . 778
8 3 6
= 2595 s
= 43 mins 15 secs Exercise 4A
6 P = (6, 11), Q = (k, 9), R = (2k, 3) Given: grad PQ = grad PR
6 11 9
k =
6 2
11 3
k 2(2k 6) = 8(k 6)
4k = 36 k = 9
7 P = (2, 3), Q = (3, 2), R = (8, z)
Since the points are collinear, their gradients are the same.
Using: grad PR = grad PQ
2 8
3
z =
2 3
3 2
z + 3 = 6(1) z = 3
8 A = (2, t), B = (7, 2t2 + 7) Given: grad AB = 2
2 7
7 2 2
t
t = 2
2t2 + 7 t = 5(2) 2t2 t 3 = 0 (t + 1)(2t 3) = 0 t = 1 or 1
2 1
9 A = (0, 6), B = (2, 1), C = (7, 3) and D = (5, 8) (i) Grad AB =
2 0
1 6
= 2
2 1
Grad BC = 2 7
1 3
=
5 2
Grad CD = 7 5
3 8
= 2
2 1
Grad DA = 0 5
6 8
=
2 1
(ii) The gradients of opp. sides are equal.
Exercise 4B
8 Given: [1(1t)]2(2t1)2 = 119t
t2 + (4t2 4t + 1) = 11 9t
5t2 + 5t 10 = 0
t2 + t 2 = 0
(t + 2)(t 1) = 0
t = 2 or 1
9 (i) AB = (51)2(22)2 = 6 units BC = (52)2 (25)2 = 18 units AC = (21)2(52)2 = 18 units
Since BC = AC, then ABC is isosceles. (Shown) (ii) Since AB is horizontal, then
area of ABC = 2
1(6)(5 2) = 9 units2
10 PQ = (33)2(41)2 = 3 units QR = (83)2(41)2 = 34 units
2
2
Since PQ2 + PR2 = QR2, by Pythagoras’ Theorem, PQR is a rt-angled .
Let the dist. from P to QR be d.
2
1(QR)(d) = 2
1(3)(5)
d = 34
15 = 2.57 units
11 Let the dist. from Q to PR be d.
Since QR is vertical, then area of PQR =
2
1(11)(5 1)
2
1(PR)(d) = 22
d = 2 2
) 3 15 ( ) 1 5 (
44
= 3.48 units
Exercise 4C
10 (i) Equation of line is y = 3 2x + c
At (3, 5), 5 =
3
2(3) + c c = 3
the equation is y = 3 2x + 3.
(ii) At (p, 3), 3 = 3 2p + 3
p = 0
11 For 2y = 5x + 7, grad. = 2 5
Equation of line is y = 2 5x + c
At (3, 2), 2 =
2
5(3) + c
c = 2 19
the equation is y = 2
19 5 x
. [or 2y = 5x 19]
12 (i) Equation of line is y = 3x + c At (3, 1), 1 = 3(3) + c
c = 8
the equation is y = 3x 8. --- (1) (ii) y = x --- (2)
(1) = (2): 3x 8 = x x = 4
14 (a) (i) Grad. of l = 0 3
3 12
= 3
(ii) Equation of l is y = 3x + c At A(0, 3), 3 = 0 + c
c = 3
the equation is y = 3x + 3.
(b) x = 3 is the line of symmetry and coordinates of B = (3, 12) So, C = (6, 3)
15 mx = ny + 2 y = n mx
n 2
Since line is parallel x-axis, then grad. = 0 m = 0 If line is parallel y-axis, then the condition required is n = 0 Exercise 6A
5 (a) x2 = 412 402 (Pythagoras’ Theorem) x = 412402 = 9 cm
(b) (i) sin PRS = sin ˆ PRQ = ˆ 9 41 (ii) cos PRS = cos ˆ PRQ = ˆ 40
41 (iii) tan PRQ = ˆ 9
40 6 Let the acute angle be x.
(c) sin x = 0.875
x = 61.04 = 61.0
(d) sin x = 0.3456
x = 20.22 = 20.2
7 Let the obtuse angle be x.
(c) sin x = 0.875
x = 180 61.04 = 119.0
(d) sin x = 0.3456
x = 180 20.22 = 159.8
9 (c) sin x = 0.4714
x = 28.13 or 180 28.13
= 28.1 or 151.9
(e) cos x = 0.783
x = 180 38.46 = 141.5
(f) cos x = 0.524 x = 58.4
12 AB = 52122 = 13 units (a) sin ABCˆ = 5
(b) cos ABCˆ = 12 13 (c) tan ACB = ˆ 5
27
Exercise 6B
4 (i) Area of triangle ABC = 2
1(32)(43) sin 67
= 633.307 633 cm2 (ii) Let h be the required perpendicular distance.
2
1(43)(h) = 633.307
h = 29.5 cm
6 (i) sin ACD = ˆ 3.7 8.0
ACD = 27.549 ˆ 27.5
(ii) cos 40.4 = 8.0 AB
AB = 10.505 10.5 cm (iii) Area of triangle AED =
2
1(3.7)(4.1) sin 55.1
6.22 cm2
8 (i) CAD = 180 90 30 ˆ = 60 (angle sum of triangle CAD) So, BADˆ = 90 60 = 30
(ii) sin 30 = 20 BD
BD = 10 cm (iii) tan 30 = 20
AC
AC = 34.641 Area of triangle ABC =
2
1(34.641)(20) 346 cm2
10 Given: 3QR = 4PR, then PR = 3 4QR Since area of triangle PQR = 158
2
1(QR)(PR) sin 55 = 158
(QR)(3
4QR) sin 55 = 316
QR 22.7 cm
Exercise 6C
6 (i) Using Sine rule, sin ˆ 7.4
BAC = sin 91 11.6
C
A
B ˆ = 39.630 39.6
(ii) ACB = 180 91 39.630 ˆ (angle sum of triangle)
= 49.370 49.4
(iii) Using Sine rule,
sin 49.370 AB
= 11.6 sin 91
AB 8.80 cm
8 (i) BAC = 180 62 68 = 50 ˆ (angle sum of triangle) Using Sine rule,
sin 62 AC
= 6 sin 50
AC = 6.916 6.92 m
(ii) BAC = 180 68 = 112 ˆ (adj, angles on a st. line) Required area =
2
1(6)(6.916) sin 68 + 2
1(6)(7.5) sin 112
= 40.1 m2 10 (i) tan ˆSPT = 5.7
4.3
SPT = 52.97 74 ˆ
By property of “alt. angles, parallel lines”, QS is not parallel to PT.
(ii) cos 63 = 4.3 PR
PR = 9.472 9.47 cm
(iii) PQS = 180 63 74 = 43 ˆ (angle sum of triangle) Using Sine rule,
sin 63 QS
= 4.3 sin 43
QS = 5.62 cm
11 QPS = 180 48 55 = 77 ˆ (angle sum of triangle) Using Sine rule,
sin 55 PQ
= 5.7 sin 77
PQ = 4.792 cos 73 =
5.7 QR
QR = 1.667 Required area =
2
1(5.7)(4.792) sin 48 + 2
1(1.667)(5.7) sin 73
= 14.7 km2 12 (i) sin 27.6 =
5.7 QR
QR = 2.641 2.64 cm
(ii) cos ˆSPR = 3.2 5.7
SPR = 55.847 55.8 ˆ (iii) Using Sine rule, sin ˆ
2.7
PST = sin 64.2 3.2
PST = 49.4 ˆ Exercise 6D
8 (i) BPAˆ = 180 60 = 120 (adj. angles on a st. line) Using Sine rule,
sin120 AB
= 5 sin 45
AB = 6.124 6.12 m
(ii) Using Cosine rule, AC2 = 52 + 82 2(5)(8) cos 60
AC = 7 m
9 (i) AMC = 180 120 = 60 (adj. angles on a st. line) ˆ Using Cosine rule, AC2 = 42 + 22 2(4)(2) cos 60
AC = 12 3.46 cm
(ii) Using Cosine rule, AB2 = 42 + 22 2(4)(2) cos 120
AB = 28 5.29 cm (iii) Using Sine rule, sin ˆ
4
ACB = sin 60 12
ACB = 90 ˆ
11 (i) Using Cosine rule, a2 = 52 + 62 2(5)(6) cos 92
a = 7.943 7.94 cm (ii) Using Cosine rule, 7.9432 = 52 + 72 2(5)(7) cos
cos = 52 72 7.9432 2(5)(7)
= 81.0
12 (i) Using Cosine rule, 4.52 = 22 + 3.52 2(2)(3.5) cos ADCˆ
cos ADCˆ =
2 2 2
2 3.5 4.5
2(2)(3.5)
ADC = 106.60 ˆ
So, ADBˆ = 180 106.60 (adj. angles on a st. line)
= 73.40 73.4
(ii) Let d be the shortest distance.
sin 73.40 = 2
d d = 1.917 1.92 cm
(iii) BADˆ = 180 50 73.40 = 56.60 (angle sum of triangle) Using Sine rule,
sin 56.60 BD
= 2 sin 50
BD = 2.18 cm
Exercise 7A
7 Let a and b be the distance from the 2 points to the foot of the clock tower, where a < b.
tan 42 = 45
a a = 49.978 tan 37 = 45
b b = 59.717
Required distance = 59.717 49.978 = 9.74 m
8 Let a and b be the heights of the mountain & mountain with castle, where a < b.
tan 45 = 55
a a = 55
tan 60 = 55
b b = 95.263
So the height of castle = 95.263 55 = 40.3 m 9 Let a be the distance from P to the foot of the bridge.
tan x = 5.5
a a = 5.5 tan x
tan 23 = 5.5 5.1
a a + 5.1 = 5.5
tan 23
5.5
tan x = 5.5
tan 23 5.1
tan x = 5.5 ÷ ( 5.5
tan 23 5.1)
x = 35.0
11 Let a and b be the distance from the 2 boats to the foot of the cliff, where a < b.
tan 23 = 88
a a = 207.32 tan 18 = 88
b b = 270.84
Required distance = 270.84 207.32 = 63.5 m Exercise 7B
5 PRQ = 135 120 = 15 (ext, angle of triangle) ˆ Using Sine rule,
sin120 PR
= 12 sin15
PR = 40.2 km
9 (a) When Nora is equidistant from the bus stop, it formed an isos. triangle at A.
Let a be the required distance.
cos 50 = 140
a a 218 m
(b) When Nora as close as possible to the bus stop,
it would be at position B, the foot of perpendicular from bus stop.
P
Q
R 135
120
12 N
N
A B 50
280
Bus stop
C
cos 50 = 280
b b 180 m
(c) When Nora is due east of the bus stop,
it would be at position C, forming a right-angled triangle.
Let c be the required distance.
cos 50 = 280
c c 436 m
10 x = 180 128 = 52 (int, angle, parallel lines) PQR = 180 52 295 ˆ = 13 (angles at a pt)
Using Cosine rule, PR2 = 252 + 302 2(25)(30) cos 13
PR 7.97 km
11 (i) Using “int, angle, parallel lines” with “angles at a pt”
PQR = (180 78) (360 305) ˆ = 47
Using Cosine rule, PR2 = 6002 + 9502 2(600)(950) cos 47
PR = 696.43 696 m (ii) Using Sine rule, sin ˆ
600
RPQ = sin 47 696.43
RPQ = 39.056 ˆ
Required bearing = 78 39.056 = 038.9
13 (i) Using Cosine rule, BC2 = 3702 + 5102 2(370)(510) cos (144 68)
BC = 552.900 553 m (ii) Using Sine rule, sin ˆ
370
ACB = sin 76 552.900
ACB = 40.49 ˆ 40.5
(iii) ABC = 180 76 40.49 = 63.51 ˆ (angle sum of triangle) Required bearing
= 360 63.51 (180 68) (angles at a pt)
= 184.5
(iv) Let d be the required distance.
sin 63.51 = 370
d d 331 m
Exercise 7C
4 (i) tan B ˆAC = 10 15 C
A
B ˆ = 56.3099 56.3
(ii) cos 56.3099 = 10 AP
AP = 5.547 5.55 m
(iii) OP2 = OA2 AP2 (Pythagoras’ Theorem) OP = 12 2 5.5472 = 10.641 10.6 m (iv) Let x be the required angle of elevation.
P
R Q 128
295
30 N
25 x
BP2 = AB2 AP2 (Pythagoras’ Theorem) BP = 10 2 5.5472 = 8.3205
tan x =
3205 . 8
641 . 10 x 52.0
5 A ˆ = 180 45 75 QB = 60 (angle sum of triangle) Using Sine rule,
75 sin
AQ =
60 sin
60
AQ = 66.921 tan 30 =
921 . 66
PQ PQ = 38.6 m
6 (i) AC2 = 202 + 202 (Pythagoras’ Theorem) AC = 800
AM = 2
1 800 = 14.142 14.1 cm (ii) OM2 = OA2 AM2 (Pythagoras’ Theorem)
OM = 2 800)2 2
(1
32 = 28.705 28.7 cm
(iii) cos O ˆAM = 32
800 5 . 0
O ˆAM = 63.8
9 (i) Let M be the midpoint of AC, which is vertically below P.
AC2 = 122 + 162 (Pythagoras’ Theorem) AC = 20
AM = 2
1(20) = 10 cm
tan P ˆAC = 10
5 C
A
P ˆ = 26.565 26.6
(ii) Using triangle PAM
PA2 = PM2 + AM2 (Pythagoras’ Theorem) PA = 5 2 102 = 125
Using Cosine rule with PA = PB, PB2 = PA2 + AB2 2(PA)(AB)cos P ˆAB cos P ˆAB =
) 12 )(
125 ( 2
) 125 ( 12 ) 125
( 2 2 2
B A
P ˆ = 57.5
11 (i) tan O ˆ = BA 30 28 A
B
O ˆ = 43.025
= 226.975 227.0
(ii) Using Cosine rule,
BC2 = OB2 + OC2 2(OB)(OC)cos C ˆOB cos C ˆOB =
) 50 )(
30 ( 2
70 50 302 2 2 B
O
C ˆ = 120
(iii) Produce BO to a point P, P
O
C ˆ = 180 120 = 60 (adj. angles on a st. line) Bearing of C from O = 270 + 60
= 330
Great angle of elevation is the shortest distance from B, and let h be the required height.
tan 29 = 30
h h = 16.6 m
13 (a) (i) A ˆ = 180 94 47 CB = 39 (angle sum of triangle) Using Sine rule,
94 sin
BC =
39 sin
240
BC = 380.435 380 m (ii) Area of triangle ABC =
2
1(240)(380.435)sin 47
= 33387.89 33400 m2 (iii) Let d be the required shortest distance.
sin 47 = 240
d d = 175.525 176 m (b) Let x be the required angle of depression.
tan x =
525 . 175
24 x 7.8