2012 - WAEC Mathematics Past Questions and Answers - page 4

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

x^o = 86 x 2(angle at centre = 2 x angle at circumference)

= 172^o

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

n = 2m

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

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

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

y = n(vertically opp. angle)

m + n + x = 180^o

< NPQ = < PQB = 50^o(alt. < s)

< PQB = 50^o

< PQR = < PQR < PQB + < QBR = 72^o

< QBR = < PQR - < PQB = 72^o - 50^o

= 22^o

< NQR = 180 - < QBR = 180^o - 22 = 158^o, the bearing of R from Q = 158^o

6 + 8 + 6 + 5 = 25