7. Let $R S T U V$ be a regular pentagon. Construct an equilateral triangle $P R S$ with point $P$ inside the pentagon. Find the measure (in degrees) of angle $P T V$.
Solution: $\square$
We have $\angle P R V=\angle S R V-\angle S R P=108^{\circ}-60^{\circ}=48^{\circ}$. Since $P R=R S=R V$, triangle $P R V$ is isosceles, so that $\angle V P R=\angle R V P=\left(180^{\circ}-\angle P R V\right) / 2=66^{\circ}$. Likewise, we have $\angle T P S=66^{\circ}$, so that
$$
\angle T P V=360^{\circ}-(\angle V P R+\angle R P S+\angle S P T)=360^{\circ}-\left(66^{\circ}+60^{\circ}+66^{\circ}\right)=168^{\circ} .
$$

Finally, by symmetry, triangle $P T V$ is isosceles $(P T=T V)$, so $\angle P T V=\angle T V P=$ $\left(180^{\circ}-\angle T P V\right) / 2=6^{\circ}$. (See the figure.)