## Problem of the Day #147: ▲
*August 13, 2011*

*Posted by Sreenath in : potd , trackback*

In triangle ABC, $\angle ABC=95^\circ$ and $\angle ACB=50^\circ$. Point P is constructed inside the triangle such that $BP=CP$ and $\frac{[BPC]}{[ABC]}=\frac{1}{5}$. Find the integer closest to $\angle BPC$.

## Comments»

no comments yet - be the first?