Trigonometric Equations Question 165
Question: If the radius of the circumcircle of an isosceles triangle $ PQR $ is equal to $ PQ(=PR), $ then the angle P is
[IIT Screening 1992; Pb. CET 2004]
Options:
A) $ \frac{\pi }{6} $
B) $ \frac{\pi }{3} $
C) $ \frac{\pi }{2} $
D) $ \frac{2\pi }{3} $
Show Answer
Answer:
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} $ .
BETA
