25. A zig-zag line starts at the point $P$, at one end of the diameter $P Q$ of a circle. Each of the angles between the zig-zag line and the diameter $P Q$ is equal to $\alpha$ as shown. After four peaks, the zig-zag line ends at the point $Q$. What is the size of angle $\alpha$ ?
A $60^{\circ}$
B $72^{\circ}$
C $75^{\circ}$
D $80^{\circ}$
E $86^{\circ}$
SolUtion
B
After four peaks the zig-zag is at the end of the diameter, so after two peaks it must be at the centre $O$ of the circle. The triangle $O P R$ is isosceles since $O P$ and $O R$ are both radii, hence angle $P R O=\alpha$ and angle $P O R=180^{\circ}-2 \alpha$. Angle $O T R=180^{\circ}-\alpha$ (angles on a straight line), and hence angle $T R O=3 \alpha-180^{\circ}$ (angles in triangle $T R O$ ).
Triangle $O T S$ is isosceles because its base angles are equal, and hence $S T=S O$. Therefore triangles $O P R$ and OTS are congruent because they both have two sides which equal the radius of the circle, with angle $180^{\circ}-2 \alpha$ between (SAS). Hence $O T=P R=T R$ (since $P R=R T$ in isosceles triangle $P R T$ ). Hence triangle $O T R$ is isosceles, and its base angles are equal, that is $180^{\circ}-2 \alpha=3 \alpha-180^{\circ}$. Solving this gives $\alpha=72^{\circ}$.