1. **Problem statement:** We have two right-angled triangles: - Triangle PQR with right angle at Q, side PQ = 2 m, angle at P = 62°, find hypotenuse PR. - Triangle JKL with right angle at K, angle at J = 30°, hypotenuse JL = 5 m, find side KL. 2. **Formulas and rules:** For right triangles, the primary trigonometric ratios are: - $ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} $ - $ \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} $ - $ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} $ We use these to find missing sides. 3. **Triangle PQR:** - Given angle $P = 62^\circ$, side adjacent to $P$ is $PQ = 2$ m. - Hypotenuse is $PR$. Using cosine: $$\cos(62^\circ) = \frac{PQ}{PR} = \frac{2}{PR}$$ Rearranged: $$PR = \frac{2}{\cos(62^\circ)}$$ Calculate $\cos(62^\circ) \approx 0.4695$: $$PR = \frac{2}{0.4695} \approx 4.26$$ Rounded to 1 significant figure: $$PR \approx 4$$ 4. **Triangle JKL:** - Given angle $J = 30^\circ$, hypotenuse $JL = 5$ m. - Side $KL$ is opposite to angle $J$. Using sine: $$\sin(30^\circ) = \frac{KL}{5}$$ Rearranged: $$KL = 5 \times \sin(30^\circ)$$ Calculate $\sin(30^\circ) = 0.5$: $$KL = 5 \times 0.5 = 2.5$$ Rounded to 2 significant figures: $$KL = 2.5$$