9. In the diagram, $P Q=P R=Q S$ and $\angle Q P R=20^{\circ}$. What is $\angle R Q S$ ?
A $50^{\circ}$
B $60^{\circ}$
C $65^{\circ}$
D $70^{\circ}$
E $75^{\circ}$
9. B Since $P Q=Q S$, triangle $P S Q$ is isosceles and hence $\angle P S Q=20^{\circ}$. Since the angles in a triangle add to $180^{\circ}$, we have $20^{\circ}+20^{\circ}+\angle S Q P=180^{\circ}$ and hence $\angle S Q P=140^{\circ}$. Since $P Q=P R$, triangle $P R Q$ is isosceles and hence $\angle P R Q=\angle R Q P$. Also $\angle P R Q+\angle R Q P+20^{\circ}=180^{\circ}$ and hence $\angle R Q P=80^{\circ}$. Since $\angle R Q S=\angle S Q P-\angle R Q P$, the size of $\angle R Q S$ is $140^{\circ}-80^{\circ}=60^{\circ}$.