24. A Consider the quadrilateral $P Q R S$ as shown with $P Q=Q R=R S, \angle R Q P=60^{\circ}$ and $\angle S R Q=100^{\circ}$.

Draw line $P R$. Since $P Q=Q R$ and $\angle P Q R=60^{\circ}$, triangle $P Q R$ is equilateral and hence $P R=P Q=Q R=R S$ and $\angle P R Q=60^{\circ}$. Since $\angle S R Q=100^{\circ}$, $\angle S R P=100^{\circ}-60^{\circ}=40^{\circ}$. Since $P R=R S$, triangle $P R S$ is isosceles and hence $\angle R P S=\angle P S R=\frac{1}{2}\left(180^{\circ}-40^{\circ}\right)=70^{\circ}$. Therefore the largest angle of the quadrilateral is $\angle Q P S=70^{\circ}+60^{\circ}=130^{\circ}$.
24. The lengths of three adjacent sides of a quadrilateral are equal. The angle between the first and second of these sides is $60^{\circ}$ and the angle between the second and third of these sides is $100^{\circ}$. What is the largest angle of the quadrilateral?
A $130^{\circ}$
B $140^{\circ}$
C $145^{\circ}$
D $150^{\circ}$
E $160^{\circ}$