In the figure PS QS RS and QSR - 100o find QPR - JAMB Mathematics 1990 Question
A
40o
B
50o
C
80o
D
100o
correct option: b
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
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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