1. **Problem statement:** Find the exact values of: a) $\sin(\angle PSR)$ b) $\tan(\angle PSR)$ c) $\sec(\angle PQR)$ Given: - Triangle PSQ is right-angled at Q with $PQ=12$ units. - Triangle SQR is right-angled at R with $SR=4$ units and $QR=3$ units. 2. **Step 1: Understand the triangles and angles** - $\angle PSR$ is an angle in triangle SQR. - $\angle PQR$ is an angle in triangle PSQ. 3. **Step 2: Find missing sides in triangle SQR** Using Pythagoras theorem in triangle SQR: $$SQ = \sqrt{SR^2 + QR^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$ 4. **Step 3: Calculate $\sin(\angle PSR)$** $\angle PSR$ is the angle at S in triangle SQR. - Opposite side to $\angle PSR$ is $QR = 3$ - Hypotenuse is $SQ = 5$ Formula: $$\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$$ So: $$\sin(\angle PSR) = \frac{3}{5}$$ 5. **Step 4: Calculate $\tan(\angle PSR)$** - Opposite side to $\angle PSR$ is $QR = 3$ - Adjacent side to $\angle PSR$ is $SR = 4$ Formula: $$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$$ So: $$\tan(\angle PSR) = \frac{3}{4}$$ 6. **Step 5: Find missing side in triangle PSQ** We know $PQ=12$ and $SQ=5$ (from step 3). Using Pythagoras theorem: $$PS = \sqrt{PQ^2 + SQ^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$$ 7. **Step 6: Calculate $\sec(\angle PQR)$** $\angle PQR$ is the angle at Q in triangle PSQ. - Adjacent side to $\angle PQR$ is $PQ = 12$ - Hypotenuse is $PS = 13$ Formula: $$\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$$ $$\sec(\theta) = \frac{1}{\cos(\theta)}$$ So: $$\cos(\angle PQR) = \frac{12}{13}$$ $$\sec(\angle PQR) = \frac{1}{\frac{12}{13}} = \frac{13}{12}$$ **Final answers:** - $\sin(\angle PSR) = \frac{3}{5}$ - $\tan(\angle PSR) = \frac{3}{4}$ - $\sec(\angle PQR) = \frac{13}{12}$