• No results found

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

N/A
N/A
Protected

Academic year: 2022

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

Copied!
18
0
0

Loading.... (view fulltext now)

Full text

(1)

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

(2)

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

(3)

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

(6)

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%

(8)

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.

Using: grad PR = grad PQ

(9)

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(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

(11)

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

(14)

(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

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

References

Related documents

q w e r t y Description Rod cover Head cover Cylinder tube Piston rod Piston Bushing Cushion valve Snap ring Tie rod Tie rod nut Wear rod Rod end nut Back up O ring Rod seal Piston

investment advice (for the relevant information requirement, see Article 24(3) of the MiFID II draft). Only then is it actually possible for banks to offer this service without

The Seckford Education Trust and our schools should be environments in which pupils and students or their parents/carers can feel comfortable and confident

However, as described in the accompanying schedule of findings and responses, we identified certain deficiencies in internal control over financial reporting that

This is the recurring motto of the unedited treatise Diez privilegios para mujeres preñadas 4 (Ten Privileges for Pregnant Women), written in 1606 by the Spanish physician

Research question: How did the experimental group (those participating in an interactive component in the hypothesis testing unit of a statistics course) differ from the control

In a surprise move, the Central Bank of Peru (BCRP) reduced its benchmark interest rate by 25 basis points (bps) to 3.25% in mid-January following disappointing economic growth data

Itron wideband systems operate across 120 channels and are designed to receive up to 120 separate transmissions simultaneously, each at 37.5 kbps (kbps = kilo bits per second)