Question

In the given figure, ABC is an isosceles triangle with $AB=AC$ and $\angle A={50}^0,\angle B$ is equals to :

• A. ${50}^0$
• B. ${65}^0$
• C. ${90}^0$
• D. ${130}^0$

Answer 1

Answer: (B)

${65}^0$

$AB=AC$

$\therefore \angle B=\angle C$ (angles opposite to equal sides are equal)

$\angle A+\angle B+\angle C={180}^{\circ }{50}^{\circ }+\angle B+\angle B={180}^{\circ }2\angle B={130}^{\circ }\Rightarrow \angle B={65}^{\circ }$