Question

In a triangle ABC, $\angle B={60}^o$ and $\angle C={80}^o$. Suppose the bisector of $\angle B$ and $\angle C$ meet at I. Find $\angle BIC.$

Answer 1

Bisector of $\angle B$ divides the angle into two equal angles

$\angle B={30}^0$

Similarly, $\angle C={40}^0$

For $△BIC$

$\angle IBC={30}^0$

$\angle ICB={40}^0$

The sum of interior angles of a triangle is 180

$\angle BIC+30+40={180}^0$

$\angle BIC={110}^0$