In a right‑angled triangle $PQR$, $\angle Q=90^{\circ }$. If $\angle P=35^{\circ }$ and the hypotenuse $PR=15\text{ cm}$, find the lengths of the other two sides $PQ$ (adjacent to $\angle P$) and $QR$ (opposite $\angle P$). Give your answers correct to two decimal places.

Step-by-step solution

1. For $\angle P$, the side adjacent is $PQ$ and the side opposite is $QR$.
2. Use $\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}$. Hence $\cos 35^{\circ }=\frac{PQ}{15}$.
3. So $PQ=15\cos 35^{\circ }$. With $\cos 35^{\circ }\approx 0.819152$, $PQ\approx 15\times 0.819152=12.2873\text{ cm}$.
4. Round to two decimals: $PQ=12.29\text{ cm}$.
5. Use $\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}$. Hence $\sin 35^{\circ }=\frac{QR}{15}$.
6. So $QR=15\sin 35^{\circ }$. With $\sin 35^{\circ }\approx 0.573576$, $QR\approx 15\times 0.573576=8.6036\text{ cm}$.
7. Round to two decimals: $QR=8.60\text{ cm}$.

Concept tested: Using sine and cosine to find unknown sides

Published 25 August 2026 · How this solution was prepared and checked

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