Problem 505

Solution 1

Since $\angle P Q S$ is an exterior angle of $\triangle Q R S$, then $\angle P Q S=\angle Q R S+\angle Q S R$, so $136^{\circ}=x^{\circ}+64^{\circ}$ or $x=136-64=72$.

## Solution 2

Since $\angle P Q S=136^{\circ}$, then $\angle R Q S=180^{\circ}-\angle P Q S=180^{\circ}-136^{\circ}=44^{\circ}$.

Since the sum of the angles in $\triangle Q R S$ is $180^{\circ}$, then $44^{\circ}+64^{\circ}+x^{\circ}=180^{\circ}$ or $x=180-44-64=72$.

ANSWER: (A)