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JAMB - Mathematics (1990 - No. 40)

In the figure, PS = QS = RS and QSR - 100o, find QPR
40o
50o
80o
100o

解释

Since PS = QS = RS

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

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

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