1. **Problem statement:** We need to find the height of a tree given two points A and B on the ground, 30 m apart, with a right angle (90°) between them at the tree's base. The angles of elevation to the top of the tree from A and B are 41° and 52°, respectively. 2. **Setup and notation:** Let the tree's base be point C, and the height of the tree be $h$. Points A and B form a right angle at C, so triangle ABC is right-angled at C with $AB=30$ m. 3. **Using trigonometry:** From points A and B, the angles of elevation to the top of the tree are given. Let $AC = x$ and $BC = 30 - x$ since $AB=30$ m. 4. **Express height $h$ in terms of $x$:** - From point A: $\tan(41^\circ) = \frac{h}{x} \implies h = x \tan(41^\circ)$ - From point B: $\tan(52^\circ) = \frac{h}{30 - x} \implies h = (30 - x) \tan(52^\circ)$ 5. **Equate the two expressions for $h$:** $$ x \tan(41^\circ) = (30 - x) \tan(52^\circ) $$ 6. **Solve for $x$:** $$ x \tan(41^\circ) = 30 \tan(52^\circ) - x \tan(52^\circ) $$ $$ x (\tan(41^\circ) + \tan(52^\circ)) = 30 \tan(52^\circ) $$ $$ x = \frac{30 \tan(52^\circ)}{\tan(41^\circ) + \tan(52^\circ)} $$ 7. **Calculate values:** - $\tan(41^\circ) \approx 0.8693$ - $\tan(52^\circ) \approx 1.2799$ $$ x = \frac{30 \times 1.2799}{0.8693 + 1.2799} = \frac{38.397}{2.1492} \approx 17.87 \text{ m} $$ 8. **Find height $h$:** $$ h = x \tan(41^\circ) = 17.87 \times 0.8693 \approx 15.53 \text{ m} $$ 9. **Final answer:** The height of the tree is approximately **16 meters** to the nearest meter.