Ex. 24. The extensions of the angle bisectors at vertices $P$ and $Q$ of triangle $P Q R$ intersect the circumscribed circle at points $P^{\prime}$ and $Q^{\prime}$, respectively. Find $P^{\prime} Q^{\prime}$, if $P Q=6$, and the radius of the circumscribed circle is 5.
Ex. 24. Answer: $P^{\prime} Q^{\prime}=3 \sqrt{10}$ or $\sqrt{10}$.
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Solution. Let the angle $P R Q$ be denoted by $\gamma$. By the Law of Sines, $\sin \gamma=\frac{6}{2 R}=\frac{3}{5}$. Then $\cos \gamma$ is either $\frac{4}{5}$ or $-\frac{4}{5}$ if the angle $\gamma$ is obtuse. Note that in either case, the chord $P^{\prime} Q^{\prime}$ connects the endpoints of an arc of length $\pi-\gamma$ and, therefore, its length is $2 R \sin \frac{\pi-\gamma}{2}=2 R \cos \frac{\gamma}{2}=10 \sqrt{\frac{1+\cos \gamma}{2}}$. Substituting the two values of $\cos \gamma$, we get two possible answers.