A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{3}\]
C) \[\frac{\pi }{2}\]
D) \[\frac{2\pi }{3}\]
Correct Answer: D
Solution :
In \[\Delta PQR,\] the radius of circumcircle is \[PQ=PR\] \[\therefore \] \[PQ=PR=\frac{PQ}{2\sin R}=\frac{QR}{2\sin P}=\frac{PR}{2\sin Q}\] \[\Rightarrow \]\[\sin R=\sin Q=\frac{1}{2}\Rightarrow \angle R=\angle Q=\frac{\pi }{6}\] \[\Rightarrow \] \[\angle P=\pi -\angle R-\angle Q=\frac{2\pi }{3}\].
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