Question

In $△ABC,AB=BC$ and $\angle B={64}^o$. Find $\angle C$.

Answer 1

We can see that $\angle A$ = $\angle C$ [$\because AB=AC$]
$\angle A$ +$\angle C$ + $\angle B$ = ${180}^{\circ }$ {$\because$ sum of all the angles of triangle is ${180}^o$}
so, $2\angle C$ = ${180}^{\circ }$ - ${64}^{\circ }$= ${116}^{\circ }$ (since $\angle A$ = $\angle C$)
$\Rightarrow \angle C$ = $\frac{{116}^{\circ }}{2}$ =${58}^{\circ }$
$\Rightarrow \angle A$ = ${58}^{\circ }$ and $\angle C$ = ${58}^{\circ }$
$\Rightarrow \angle C$ = ${58}^{\circ }$