1. The problem states that triangles $\triangle QRP$ and $\triangle STP$ are similar, denoted as $\triangle QRP \sim \triangle STP$. 2. Similar triangles have corresponding angles equal and corresponding sides proportional. 3. Since both triangles share vertex $P$ and have a right angle at $P$, the angles at $P$ are equal. 4. The corresponding vertices are matched as $Q \leftrightarrow S$, $R \leftrightarrow T$, and $P \leftrightarrow P$. 5. We analyze each trigonometric statement: - $\cos R = \frac{PS}{PT}$: Angle $R$ in $\triangle QRP$ corresponds to angle $T$ in $\triangle STP$. $\cos R$ is adjacent over hypotenuse in $\triangle QRP$, which is $PR/QR$. $PS/PT$ is a ratio of sides in $\triangle STP$, but $PS$ and $PT$ correspond to sides opposite and hypotenuse of angle $T$, so this is not correct. - $\tan Q = \frac{PS}{PT}$: Angle $Q$ in $\triangle QRP$ corresponds to angle $S$ in $\triangle STP$. $\tan Q$ is opposite over adjacent in $\triangle QRP$, which is $PR/PQ$. $PS/PT$ is opposite over hypotenuse in $\triangle STP$, so this is not the tangent ratio. - $\sin S = \frac{PR}{QR}$: Angle $S$ in $\triangle STP$ corresponds to angle $Q$ in $\triangle QRP$. $\sin S$ is opposite over hypotenuse in $\triangle STP$, which corresponds to $PR/QR$ in $\triangle QRP$. This matches the sine ratio correctly. - $\tan T = \frac{PR}{PQ}$: Angle $T$ in $\triangle STP$ corresponds to angle $R$ in $\triangle QRP$. $\tan T$ is opposite over adjacent in $\triangle STP$, but $PR/PQ$ are sides in $\triangle QRP$, so this is not correct. 6. Therefore, the true statement is: $$\sin S = \frac{PR}{QR}$$