JAMB - Mathematics (1990 - No. 40)

In the figure, PS = QS = RS and QSR - 100^o, find QPR

40^o

50^o

80^o

100^o

Since PS = QS = RS

S is the centre of circle passing through P, Q, R /PS/ = /RS/ = radius

QPR \(\pm\) \(\frac{100^o}{2}\) = 50^o