Question

AB=AC and $\angle B={58}^0$, then find the value of $\angle A$.

Answer 1

We have,$AB=AC$

$\angle B={58}^{\circ }$

Since, $ABC$ is an isosceles triangle.

So, $\angle B=\angle C={58}^{\circ }$

We know that

$\angle A+\angle B+\angle C={180}^{\circ }$

$\angle A+{58}^{\circ }+{58}^{\circ }={180}^{\circ }$

$\angle A={180}^{\circ }-{116}^{\circ }$

$\angle A={64}^{\circ }$

Hence, the value of $\angle A$ is ${64}^{\circ }$.