Question

Consider isosceles triangle $ABC$, in which $AB=AC$ and$\angle ABC={50}^o$. Find the $\angle BAC$.

• A. ${50}^o$
• B. ${65}^o$
• C. ${80}^o$
• D. ${130}^o$

Answer 1

The correct option is C ${80}^o$

Here, $△ABC$ is an isosceles triangle.

Given, $AB=AC$ and $\angle ABC={50}^o$.

We know, by isosceles triangle property, angles opposite to equal sides are equal.

Thus, $\angle ABC=\angle ACB={50}^o$.

Therefore, by angle sum property of triangle,

$\angle BAC+\angle ABC+\angle ACB={180}^o$

$⟹$ $\angle BAC+{50}^o+{50}^o={180}^o$

$⟹$ $\angle BAC={180}^o-({50}^o+{50}^o)={80}^o$.

Hence, $\angle BAC={80}^o$.