In a quadrilateral $ABCD$, $\angle A : \angle B : \angle C : \angle D = 2:3:4:6$. Then $\angle B$ equals:

In a quadrilateral $ABCD$, $\angle A : \angle B : \angle C : \angle D = 2:3:4:6$. Then $\angle B$ equals:
  1. $48^{\circ}$
  2. $72^{\circ}$
  3. $96^{\circ}$
  4. $144^{\circ}$

Solution

Sum $= 15x = 360 \Rightarrow x = 24^{\circ}$. $\angle B = 3x = 72^{\circ}$.

Asked in: IMO

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