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WAEC Mathematics Past Questions & Answers - Page 133

661.

In the diagram, angle 20^o is subtended at the centre of the circle, find the value of x

30^o

50^o

80^o

100^o

View answer & explanation

Correct answer is B

2(30 + x) = 360 - 200

60 + 2x = 160

2x = 160 - 60

2x = 100

x = $\frac{100}{2}$

x = 50

662.

If /XY/ = 50m, how far cast of X is Y?

25.5m

40.6m

40.8m

43.3m

View answer & explanation

Correct answer is D

Required distance = x

where $\frac{1}{50m}$ = sin 60 x 50m x 0.866

x = 43.3m

663.

Using the diagram, find the bearing of X from Y

300^o

240^o

120^o

60^o

View answer & explanation

Correct answer is B

The bearing of X from Y

= 360 - 120 = 240

664.

In the diagram, \SQ\ = 4cm, \PT\ = 7cm. /TR/ = 5cm and ST//OR. If /SP/ = xcm, find the value of x

5.6

6.5

6.6

6.8

View answer & explanation

Correct answer is A

$\frac{/PS/}{/PQ/} = \frac{/PT/}{/PR/}$

$\frac{x}{4 + x} = \frac{7}{7 + 5}$

$\frac{x}{4 + x} = \frac{7}{12}$

(2x = 7)(4 + x); 12x = 28 + 7x

12x - 7x = 28

5x = 28

x = $\frac{28}{5}$

x = 5.6

665.

In the diagram, IG is parallel to JE, JEF = 120^o and FHG = 130^o, fins the angle marked t

40^o

70^o

80^o

100^o

View answer & explanation

Correct answer is B

Now, t + 60^o + 50^o = 180^o

t = 180^o - 110^o

t = 70^o

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