In the diagram 0 is the centre of the circle Fi... - WAEC Mathematics 2010 Question

POQ in a straight line

Hence, < POQ + < QOR = 180^o

56^o + < QOR = 180^o

< QOR = 180^o - 56^o

= 124^o

Now, in \(\bigtriangleup\) QOR OR = OQ = Radius

< ORQ = < OQR = 2x (Base angles of an Isosceles \(\bigtriangleup\))

2x + 124 + 2x = 180^o

4x + 124 = 180

4x = 180 - 124

4x = 56

x = \(\frac{56}{4}\)

x = 14^o