In the diagram, points $P_{1}, P_{3}, P_{5}, P_{7}$ are on $B A$ and points $P_{2}, P_{4}, P_{6}, P_{8}$ are on $B C$ such that $B P_{1}=P_{1} P_{2}=P_{2} P_{3}=P_{3} P_{4}=P_{4} P_{5}=P_{5} P_{6}=P_{6} P_{7}=P_{7} P_{8}$. If $\angle A B C=5^{\circ}$, what is the measure of $\angle A P_{7} P_{8}$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_56af4dba662dd63eccc5g-1.jpg?height=225&width=873&top_left_y=1606&top_left_x=672)
Since $B P_{1}=P_{1} P_{2}=P_{2} P_{3}=P_{3} P_{4}=P_{4} P_{5}=P_{5} P_{6}=P_{6} P_{7}=P_{7} P_{8}$, then each of $\triangle B P_{1} P_{2}$, $\triangle P_{1} P_{2} P_{3}, \triangle P_{2} P_{3} P_{4}, \triangle P_{3} P_{4} P_{5}, \triangle P_{4} P_{5} P_{6}, \triangle P_{5} P_{6} P_{7}$, and $\triangle P_{6} P_{7} P_{8}$ is isosceles.

Since $\angle A B C=5^{\circ}$, then $\angle B P_{2} P_{1}=\angle A B C=5^{\circ}$.

Next, $\angle P_{2} P_{1} P_{3}$ is an exterior angle for $\triangle B P_{1} P_{2}$.

Thus, $\angle P_{2} P_{1} P_{3}=\angle P_{1} B P_{2}+\angle P_{1} P_{2} B=10^{\circ}$.

(To see this in another way, $\angle B P_{1} P_{2}=180^{\circ}-\angle P_{1} B P_{2}-\angle P_{1} P_{2} B$ and

$$
\angle P_{2} P_{1} P_{3}=180^{\circ}-\angle B P_{1} P_{2}=180^{\circ}-\left(180^{\circ}-\angle P_{1} B P_{2}-\angle P_{1} P_{2} B\right)=\angle P_{1} B P_{2}+\angle P_{1} P_{2} B
$$

The first of these equations comes from the sum of the angles in the triangle and the second from supplementary angles.)

Continuing in this way,

$$
\begin{aligned}
& \angle P_{2} P_{3} P_{1}=\angle P_{2} P_{1} P_{3}=10^{\circ} \\
& \angle P_{3} P_{2} P_{4}=\angle P_{2} P_{3} P_{1}+\angle P_{2} B P_{3}=15^{\circ} \\
& \angle P_{3} P_{4} P_{2}=\angle P_{3} P_{2} P_{4}=15^{\circ} \\
& \angle P_{4} P_{3} P_{5}=\angle P_{3} P_{4} P_{2}+\angle P_{3} B P_{4}=20^{\circ} \\
& \angle P_{4} P_{5} P_{3}=\angle P_{4} P_{3} P_{5}=20^{\circ} \\
& \angle P_{5} P_{4} P_{6}=\angle P_{4} P_{5} P_{3}+\angle P_{4} B P_{5}=25^{\circ} \\
& \angle P_{5} P_{6} P_{4}=\angle P_{5} P_{4} P_{6}=25^{\circ} \\
& \angle P_{6} P_{5} P_{7}=\angle P_{5} P_{6} P_{4}+\angle P_{5} B P_{6}=30^{\circ} \\
& \angle P_{6} P_{7} P_{5}=\angle P_{6} P_{5} P_{7}=30^{\circ} \\
& \angle P_{7} P_{6} P_{8}=\angle P_{6} P_{7} P_{5}+\angle P_{6} B P_{7}=35^{\circ} \\
& \angle P_{7} P_{8} P_{6}=\angle P_{7} P_{6} P_{8}=35^{\circ} \\
& \angle A P_{7} P_{8}=\angle P_{7} P_{8} P_{6}+\angle P_{7} B P_{8}=40^{\circ}
\end{aligned}
$$