1. **State the problem:** We have a right triangle PQR with $\angle R = 90^\circ$, $\angle P = 24^\circ$, and hypotenuse $PQ = 7.5$ cm. We need to find the lengths of sides $RQ$ and $PR$ to the nearest tenth of a centimetre. 2. **Recall the trigonometric ratios:** - $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$ - $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$ - $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$ 3. **Identify sides relative to angle $P$:** - Opposite side to $\angle P$ is $RQ$ - Adjacent side to $\angle P$ is $PR$ - Hypotenuse is $PQ = 7.5$ cm 4. **Calculate $RQ$ using sine:** $$RQ = PQ \times \sin(24^\circ) = 7.5 \times \sin(24^\circ)$$ Calculate $\sin(24^\circ) \approx 0.4067$: $$RQ = 7.5 \times 0.4067 = 3.05025$$ Rounded to nearest tenth: $$RQ \approx 3.1 \text{ cm}$$ 5. **Calculate $PR$ using cosine:** $$PR = PQ \times \cos(24^\circ) = 7.5 \times \cos(24^\circ)$$ Calculate $\cos(24^\circ) \approx 0.9135$: $$PR = 7.5 \times 0.9135 = 6.85125$$ Rounded to nearest tenth: $$PR \approx 6.9 \text{ cm}$$ 6. **Final answer:** The lengths of the sides are: - $RQ \approx 3.1$ cm - $PR \approx 6.9$ cm