In isosceles $\triangle ABC$ with $AB = AC$ and $\angle B = 65^{\circ}$, $\angle A$ equals
In isosceles $\triangle ABC$ with $AB = AC$ and $\angle B = 65^{\circ}$, $\angle A$ equals
- $50^{\circ}$
- $65^{\circ}$
- $130^{\circ}$
- $115^{\circ}$
Solution
$\angle B = \angle C = 65^{\circ}$. $\angle A = 180 - 130 = 50^{\circ}$.
Asked in: MH-SSC-9
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