WAEC Past Questions and Answers - Page 2919

14,591.

In the diagram, MQ//RS, < TUV = 70^o and < RLV = 30^o. Find the value of x

150^o

110^o

100^o

95^o

View answer & explanation

Correct answer is C

L + 30^o - 180^o(Sum of < s on straight line)

L = 180^o - 30^o = 150^o

L = q = 150^o(opposite < s are equal)

y = b = 30^o(alt. < s)

b + c = 180^o(sum of < s on str. line)

30^o + c 180

c = 180 - 30

c = 150^o

b = a = 30^o (opp < s are equal)

c = d = 150^o (opp < s are equal)

a + k + 70^o = 180^o (sum of < s on \(\bigtriangleup\))

30^o + k + 70^o = 180

k + 100^o = 180

k = 180 - 100

k = 80^o

x + 80^o = 180(sum of < s on straight line)

x = 180^o - 80^o

x = 100^o

14,592.

The diagram is a circle centre O. If < SPR = 2m and < SQR = n, express m in terms of n

m = \(\frac{n}{2}\)

m = 2n

m = n - 2

m = n + 2

View answer & explanation

Correct answer is A

If < SPR = 2m then < SQR = 2m but < AQR was n

n = 2m

m = \(\frac{n}{2}\)

14,593.

In the diagram, O is a circle centre of the circle PQRS and < PSR = 86^o. If < PQR = x^o, find x

108^o

172^o

130^o

50^o

View answer & explanation

Correct answer is B

No explanation has been provided for this answer.

14,594.

In the diagram, |QR| = 10m, |SR| = 8m

< QPS = 30^o, < QRP = 90^o and |PS| = x, Find x

1.32m

6.32m

9.32m

17.32

View answer & explanation

Correct answer is C

In right angled \(\bigtriangleup\)QPR

tan 30^o = \(\frac{10}{x + 8}\)

(x + 8) tan 30 = 10

x + 8 = \(\frac{10}{0.5773}\)

x +8 = 17.3

x = 17.3 - 8

x = 9.32