2012 - WAEC Mathematics Past Questions and Answers - page 4
< QPS = 30o, < QRP = 90o and |PS| = x, Find x
In right angled (\bigtriangleup)QPR
tan 30o = (\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
xo = 86 x 2(angle at centre = 2 x angle at circumference)
= 172o
If < SPR = 2m then < SQR = 2m but < AQR was n
n = 2m
m = (\frac{n}{2})
L + 30o - 180o(Sum of < s on straight line)
L = 180o - 30o = 150o
L = q = 150o(opposite < s are equal)
y = b = 30o(alt. < s)
b + c = 180o(sum of < s on str. line)
30o + c 180
c = 180 - 30
c = 150o
b = a = 30o (opp < s are equal)
c = d = 150o (opp < s are equal)
a + k + 70o = 180o (sum of < s on (\bigtriangleup))
30o + k + 70o = 180
k + 100o = 180
k = 180 - 100
k = 80o
x + 80o = 180(sum of < s on straight line)
x = 180o - 80o
x = 100o
m + y + x = 180o(sum of < s on straight line)
y = n(vertically opp. angle)
m + n + x = 180o
< NPQ = < PQB = 50o(alt. < s)
< PQB = 50o
< PQR = < PQR < PQB + < QBR = 72o
< QBR = < PQR - < PQB = 72o - 50o
= 22o
< NQR = 180 - < QBR = 180o - 22 = 158o, the bearing of R from Q = 158o