15. A In the diagram shown, $T$ is the foot of the perpendicular from $R$ to $P Q$ produced. Consider triangle $Q T R$. Angle $Q T R=90^{\circ}$ and, because the sum of angles on a straight line $=180^{\circ}$, angle $R Q T=(180-2 \times 60)^{\circ}=60^{\circ}$. So $\sin 60^{\circ}=\frac{R T}{Q R}$ and hence $R T=Q R \times \sin 60^{\circ}=2 \times \frac{\sqrt{3}}{2}=\sqrt{3}$. Also $\cos 60^{\circ}=\frac{Q T}{Q R}$, so $Q T=Q R \times \cos 60^{\circ}=2 \times \frac{1}{2}=1$. Therefore $P T=7$.
By Pythagoras' Theorem, $P R^{2}=P T^{2}+T R^{2}=7^{2}+(\sqrt{3})^{2}$ $=49+3=52$. So $P R=\sqrt{52}=\sqrt{4} \times \sqrt{13}=2 \sqrt{13}$.
15. The diagram shows six congruent equilateral triangles, of side-length 2, placed together to form a parallelogram.
What is the length of $P R$ ?

A $2 \sqrt{13}$
B 7
C $6 \sqrt{3}$
D 9
E $7 \sqrt{3}$