Question

ABC is a triangle in which AB = AC. The base BC is produced to D and $\angle ACD={130}^{\circ }$. Then $\angle A$ equals

• A. ${80}^{\circ }$
• B. ${60}^{\circ }$
• C. ${50}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

The correct option is A ${80}^{\circ }$

Given, AB = AC.
$\therefore \angle ABC=\angle ACB$ (angles opposite to equal sides of a triangle are equal)
$\angle ACD={130}^{\circ }$ (given)
$\angle ACB={180}^{\circ }-{130}^{\circ }={50}^{\circ }$ [linear pair]
$⟹\angle ABC=\angle ACB={50}^{\circ }$
$\therefore \angle A={180}^{\circ }-(\angle ABC+\angle ACB)$
$⟹\angle A={180}^{\circ }-({50}^{\circ }+{50}^{\circ })$
Hence, $\angle A={80}^{\circ }$.