In quadrilateral $ABCD$, $\angle A = \angle B$ and $\angle C = 100^{\circ}, \angle D = 80^{\circ}$. Then…

In quadrilateral $ABCD$, $\angle A = \angle B$ and $\angle C = 100^{\circ}, \angle D = 80^{\circ}$. Then $\angle A$ equals:
  1. $80^{\circ}$
  2. $90^{\circ}$
  3. $100^{\circ}$
  4. $110^{\circ}$

Solution

$2 \angle A + 100 + 80 = 360 \Rightarrow 2\angle A = 180 \Rightarrow \angle A = 90^{\circ}$.

Asked in: IMO

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