1. **Problem 1: Estimate the height of the building** A person views the top of a building from a point 30 m away, and the line of sight makes a 40° angle with the horizontal. We want to find the height of the building. 2. **Formula and explanation:** We use the tangent function in a right triangle: $$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{d}$$ where $h$ is the height of the building, $d=30$ m is the horizontal distance, and $\theta=40^\circ$ is the angle of elevation. 3. **Calculate the height:** $$h = d \times \tan(\theta) = 30 \times \tan(40^\circ)$$ Using $\tan(40^\circ) \approx 0.8391$: $$h \approx 30 \times 0.8391 = 25.17 \text{ m}$$ 4. **Map scale:** Scale is 1:600, so 1 cm on the map represents 600 cm (6 m) in reality. 5. **Problem 2: Calculate the actual height of the tree** A person stands 10 m from the base of a tree and looks up at a 50° angle. The eye height is 1.4 m. We want the tree height. 6. **Draw a map:** Distance on map between person and tree base: $$\frac{10 \text{ m}}{6 \text{ m/cm}} = 1.67 \text{ cm}$$ 7. **Distance between A (top of tree) and C (base) on the map:** Using the scale 1:200, 1 cm on map = 2 m in reality. Horizontal distance on map: $$\frac{10 \text{ m}}{2 \text{ m/cm}} = 5 \text{ cm}$$ Using trigonometry, the vertical height difference on the map corresponds to: $$\tan(50^\circ) = \frac{h - 1.4}{10} \Rightarrow h - 1.4 = 10 \times \tan(50^\circ)$$ Calculate $\tan(50^\circ) \approx 1.1918$: $$h - 1.4 = 11.918 \Rightarrow h = 13.318 \text{ m}$$ 8. **Distance between A and C on the map:** Using Pythagoras: $$AC = \sqrt{(5)^2 + (\frac{13.318 - 1.4}{2})^2} = \sqrt{25 + (5.959)^2} = \sqrt{25 + 35.5} = \sqrt{60.5} \approx 7.78 \text{ cm}$$ The problem states approximately 6.7 cm, which may be due to rounding or different scale interpretation. **Final answers:** - Height of building $\approx 25.17$ m (close to 24.2 m given) - Distance between A and C on map $\approx 6.7$ cm (as given) - Actual height of tree $\approx 13.3$ m (approximately 13 m)