Question

In a $△ABC,AB=AC$ and $\angle A={50}^o$. Find $\angle B$ and $\angle C$.

Answer 1

We can see that, $\angle B$ = $\angle C$ (Since AB = AC)
$\angle B$ + $\angle C$ +$\angle A$ = ${180}^{\circ }$ [sum of all the angles of a triangle is ${180}^o$]
$\Rightarrow \angle B$ + $\angle C$ = ${180}^{\circ }-{50}^{\circ }$
$\Rightarrow$ $2\angle C$ = ${180}^{\circ }-{50}^{\circ }$
$\Rightarrow \angle C$ =$\frac{{130}^{\circ }}{2}$
$\Rightarrow \angle B$ = ${65}^{\circ }$ and $\angle C$ = ${65}^{\circ }$