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
  1. $50^{\circ}$
  2. $65^{\circ}$
  3. $130^{\circ}$
  4. $115^{\circ}$

Solution

$\angle B = \angle C = 65^{\circ}$. $\angle A = 180 - 130 = 50^{\circ}$.

Asked in: MH-SSC-9

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