# Semiperimeter

Author: karavpetr
Problem has been solved: 42 times

Русский язык | English Language

For the triangle $ABC$ the following property holds : $BC \cdot AC \cdot (1 - \sin \angle ACB) = 2 p (p - AB)$, where $p$ is the semiperimeter of the triangle $ABC$. Find $\angle ACB$ in degrees.