In parallelogram $PQRS$, $\angle P = 70^{\circ}$. Then $\angle Q$ equals:

In parallelogram $PQRS$, $\angle P = 70^{\circ}$. Then $\angle Q$ equals:
  1. $70^{\circ}$
  2. $110^{\circ}$
  3. $90^{\circ}$
  4. $140^{\circ}$

Solution

$\angle P$ and $\angle Q$ are adjacent angles of a parallelogram, so they are supplementary: $\angle Q = 180 - 70 = 110^{\circ}$.

Asked in: IMO

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