25. Two chords $P Q$ and $P R$ are drawn in a circle with diameter $P S$. The point $T$ lies on $P R$ and $Q T$ is perpendicular to $P R$. The angle $Q P R=60^{\circ}, P Q=24 \mathrm{~cm}, R T=3 \mathrm{~cm}$. What is the length of the chord $Q S$ in $\mathrm{cm}$ ?
A $\sqrt{3}$
B 2
C 3
D $2 \sqrt{3}$
E $3 \sqrt{2}$
25. D $P S$ is a diameter, so $\angle P R S=90^{\circ}$ (angle in a semicircle). Let $U$ be the point on $Q T$ for which $S U$ is perpendicular to $Q T$. Hence RSUT is a rectangle, and $S U=T R=3$.
In triangle $T P Q$, we have $\angle P T Q=90^{\circ}$ and $\angle Q P T=60^{\circ}$, so $\angle T Q P=30^{\circ}$. Also, $\angle P Q S=90^{\circ}$ (angle in a semicircle) so $\angle U Q S=90^{\circ}-\angle T Q P=90^{\circ}-30^{\circ}=60^{\circ}$.
Also $\sin \angle U Q S=\frac{U S}{Q S}=\frac{3}{Q S}$. So
$$
Q S=\frac{3}{\sin \angle U Q S}=\frac{3}{\frac{1}{2} \sqrt{3}}=3 \times \frac{2}{\sqrt{3}}=2 \sqrt{3} .
$$