# Quadratic Equations & Functions Exercise 1A 4 (f) x 2 = 0. x 2 ) 2 = ) 2 ( = ± 7. x = 0.81 or 0.05

## Full text

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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

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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

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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)

(4)

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

(5)

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

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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) = 2g1h =  g

h 2

8 (c)

4

2

3 3



 

f

e  

 

11

27 9

f

e = 8

81 12

f

e9

11

27e

f = 3e3f 3

(7)

(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 = 102d = 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) (e3f 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

)45 32m4n8 = m12n1  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%

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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  102

= 1.53  101

(h) 6 8

10

10 5 . 3 10 2 . 7

10 4 . 2

 = 0.0033496  102

= 3.35  105 7 (d) 3 9.261106 = 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  103, y = 7  104

x + 8y = 2  103 + 8(7  104) = 7.6  103 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.

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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

1 3

 =

5 2

3 8

 = 2

2 1

6 8

 =

2 1

(ii) The gradients of opp. sides are equal.

Exercise 4B

8 Given: [1(1t)]2(2t1)2 = 119t

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 = (51)2(22)2 = 6 units BC = (52)2 (25)2 = 18 units AC = (21)2(52)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 = (33)2(41)2 = 3 units QR = (83)2(41)2 = 34 units

2

2 

(10)

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

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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 = 412402 = 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 = 52122 = 13 units (a) sin ABCˆ = 5

(12)

(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

(13)

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

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(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

2 2 2

2 3.5 4.5

2(2)(3.5)

 

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

dd = 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

(15)

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

aa = 49.978 tan 37 = 45

bb = 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

aa = 55

tan 60 = 55

bb = 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

aa = 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

aa = 207.32 tan 18 = 88

bb = 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

aa  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

(16)

cos 50 = 280

bb  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

cc  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

dd  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

(17)

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

PQPQ = 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

( 222

B A

P ˆ = 57.5

11 (i) tan O ˆ = BA 30 28 A

B

O ˆ = 43.025

(18)

= 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 30222 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

hh = 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

dd = 175.525  176 m (b) Let x be the required angle of depression.

tan x =

525 . 175

24 x  7.8

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