Setup: Let the two observers stand on the ground. Let the basket be at height $h$ m. Let the horizontal distance from the first observer (60°) to the point directly below the basket be $d$ m. The second observer is 100 m farther away, so their horizontal distance is $(d + 100)$ m.
From first observer:
$$\tan 60° = \frac{h}{d} \Rightarrow \sqrt{3} = \frac{h}{d} \Rightarrow d = \frac{h}{\sqrt{3}} \quad \cdots(1)$$
From second observer:
$$\tan 30° = \frac{h}{d+100} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{d+100} \Rightarrow d+100 = h\sqrt{3} \quad \cdots(2)$$
Substituting (1) into (2):
$$\frac{h}{\sqrt{3}} + 100 = h\sqrt{3}$$
$$100 = h\sqrt{3} - \frac{h}{\sqrt{3}} = h\left(\frac{3-1}{\sqrt{3}}\right) = \frac{2h}{\sqrt{3}}$$
$$h = \frac{100\sqrt{3}}{2} = 50\sqrt{3} \text{ m}$$
(a) Height of the basket:
$$h = 50\sqrt{3} \approx 86.6 \text{ m}$$
(b) Distance of basket from first observer's eye (line of sight):
$$d = \frac{50\sqrt{3}}{\sqrt{3}} = 50 \text{ m}$$
$$\text{Line of sight} = \frac{h}{\sin 60°} = \frac{50\sqrt{3}}{\frac{\sqrt{3}}{2}} = 100 \text{ m}$$
(c) Horizontal distance of second observer from the basket:
$$d + 100 = 50 + 100 = 150 \text{ m}$$
Source: Chapter 9 — Heights and Distances, Section 9.1
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