Question

In the adjoining figure AB = AC and $\angle ACD={110}^{\circ }$. Find $\angle A$.

Answer 1

We have
AB=AC
$\Rightarrow \angle B=\angle C$
Now $\angle ACB+\angle ACD={180}^{\circ }$
(Angles of a linear pair)
$\Rightarrow \angle C+{110}^{\circ }={180}^{\circ }$
$\Rightarrow \angle C={70}^{\circ }$
$\therefore \angle B={70}^{\circ }$
($\because$ triangles is an isosceles triangle )
Now $\angle A+\angle B+\angle C={180}^{\circ }$
$\therefore \angle B=\angle C$
$\Rightarrow \angle A+{70}^{\circ }+{70}^{\circ }={180}^{\circ }$
$\Rightarrow \angle A={180}^{\circ }-{140}^{\circ }$
$\Rightarrow \angle A={40}^{\circ }$