1. **State the problem:** A student stands at point A near a cliff and walks 40 m back to point B. The angle of elevation to the top of the cliff from A is 49° and from B is 32°. We need to find the height of the cliff. 2. **Set up variables and diagram:** Let $h$ be the height of the cliff and $x$ be the horizontal distance from point A to the base of the cliff. 3. **Use tangent of angles:** From point A, $$\tan(49^\circ) = \frac{h}{x} \implies h = x \tan(49^\circ)$$ From point B, which is 40 m behind A, the horizontal distance to the cliff is $x + 40$, so: $$\tan(32^\circ) = \frac{h}{x + 40} \implies h = (x + 40) \tan(32^\circ)$$ 4. **Set the two expressions for $h$ equal:** $$x \tan(49^\circ) = (x + 40) \tan(32^\circ)$$ 5. **Solve for $x$:** $$x \tan(49^\circ) = x \tan(32^\circ) + 40 \tan(32^\circ)$$ $$x \tan(49^\circ) - x \tan(32^\circ) = 40 \tan(32^\circ)$$ $$x (\tan(49^\circ) - \tan(32^\circ)) = 40 \tan(32^\circ)$$ $$x = \frac{40 \tan(32^\circ)}{\tan(49^\circ) - \tan(32^\circ)}$$ 6. **Calculate values:** $$\tan(49^\circ) \approx 1.1504$$ $$\tan(32^\circ) \approx 0.6249$$ So, $$x = \frac{40 \times 0.6249}{1.1504 - 0.6249} = \frac{24.996}{0.5255} \approx 47.57$$ 7. **Find height $h$:** $$h = x \tan(49^\circ) = 47.57 \times 1.1504 \approx 54.7$$ **Final answer:** The height of the cliff is approximately **54.7 metres**.