Question

In a $\Delta ABC,ifAB=AC\text{and BC is produced}$

$\text{to D such that}\angle ACD={100}^{\circ },then\angle A=$

• A. ${20}^{\circ }$
• B. ${40}^{\circ }$
• C. ${60}^{\circ }$
• D. ${80}^{\circ }$

Answer 1

The correct option is A

${20}^{\circ }$

(a),

In $\Delta ABC,AB=AC$

$\therefore \angle B=\angle C$

$ButExt.\angle ACD=\angle A+\angle B$

$\angle ACB+\angle ACD={180}^{\circ }\text{(Linear pair)}$

$\therefore \angle ACB+{100}^{\circ }={180}^{\circ }$

$\Rightarrow \angle ACB={180}^{\circ }-{100}^{\circ }={80}^{\circ }$

$\therefore \angle B=\angle ACD={80}^{\circ }$

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

$\therefore \angle A+{80}^{\circ }+{80}^{\circ }={180}^{\circ }$

$\Rightarrow \angle A+{160}^{\circ }={180}^{\circ }$

$\therefore \angle A={180}^{\circ }-{160}^{\circ }={20}^{\circ }$