Question

In the fig $AB=AC$ and $\angle ACD=120$ find $\angle A$.

Answer 1

$\because AB=AC\Rightarrow \angle ABC=\angle ACB=x$(say)

Let $\angle BAC=y$

We know,

Exterior angles $=$ sun of interior opposite angles

or, ${120}^o=\angle ABC+\angle BAC$

or, ${120}^o=x+y---\left(1\right)$

Again, $\angle ACB+\angle ACD={180}^o$

or, $x+{120}^o={180}^o$

$\therefore x={60}^o$

From $\left(1\right)$,

${60}^o+y={120}^o\Rightarrow y={60}^o\therefore \angle A={60}^o$