1. The problem asks to find missing angles and heights in right-angled triangles using trigonometric ratios. 2. For angles opposite a side, use sine: $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$. 3. For heights opposite an angle, use $\sin \theta = \frac{h}{\text{hypotenuse}}$ and solve for $h$. 4. (a) Given $\sin w = \frac{8}{10} = 0.8$, find $w$ by $w = \sin^{-1}(0.8)$. 5. Calculate $w$: $w = 53.13^\circ$ (approx). 6. (b) For angle $x$, opposite side 15 cm, hypotenuse 12 cm is impossible since hypotenuse must be longest side. Likely sides swapped; assume hypotenuse 15 cm, opposite 12 cm. 7. Calculate $x$: $\sin x = \frac{12}{15} = 0.8$, so $x = \sin^{-1}(0.8) = 53.13^\circ$. 8. (c) For angle $y$, opposite side 14 cm, hypotenuse 9 cm is impossible; hypotenuse must be longest side. Assuming hypotenuse 14 cm, opposite 9 cm. 9. Calculate $y$: $\sin y = \frac{9}{14} \approx 0.6429$, so $y = \sin^{-1}(0.6429) = 40.06^\circ$. 10. (4a) Calculate height $h$ opposite $40^\circ$ with hypotenuse 12 cm: $$\sin 40^\circ = \frac{h}{12} \Rightarrow h = 12 \times \sin 40^\circ$$ 11. Calculate $h$: $h = 12 \times 0.6428 = 7.71$ cm. 12. (4b) Calculate height $h$ opposite $37^\circ$ with hypotenuse 17 cm: $$\sin 37^\circ = \frac{h}{17} \Rightarrow h = 17 \times \sin 37^\circ$$ 13. Calculate $h$: $h = 17 \times 0.6018 = 10.23$ cm. 14. (4c) Calculate height $h$ adjacent to $29^\circ$ with base 15 cm. Use cosine: $$\cos 29^\circ = \frac{h}{15} \Rightarrow h = 15 \times \cos 29^\circ$$ 15. Calculate $h$: $h = 15 \times 0.8746 = 13.12$ cm. Final answers: - (a) $w = 53.13^\circ$ - (b) $x = 53.13^\circ$ - (c) $y = 40.06^\circ$ - (4a) $h = 7.71$ cm - (4b) $h = 10.23$ cm - (4c) $h = 13.12$ cm