Question

In a triangle ABC, AB=AC. The base BC is produced to D and $\angle ACD={130}^0$. Then $\angle A$ equals

• A. ${80}^0$
• B. ${60}^0$
• C. ${50}^0$
• D. ${40}^0$

Answer 1

Answer: (A)

${80}^0$

Given: $AB=AC$, $\angle ACD=130$
$\angle ACD+\angle ACB=180$ (Linear pair)
$130+\angle ACB=180$
$\angle ACB={50}^{\circ }$
Now, In $△ABC$,
$\angle ABC=\angle ACB=50$ (Isosceles triangle property)
Sum of angles = 180
$\angle ABC+\angle ACB+\angle BAC=180$
$50+50+\angle BAC=180$
$\angle BAC={80}^{\circ }$