Question: From the top of a vertical cliff, two boats are seen on opposite sides of the cliff, in line with its base. Their angles of depression are $30^\circ$ and $60^\circ$, and the boats are 200 m apart. Find the height of the cliff and the distance of each boat from its foot.

1. **State the problem:** We have a vertical cliff with two boats on opposite sides. The angles of depression to the boats are $30^\circ$ and $60^\circ$, and the distance between the boats is 200 m. We need to find the height of the cliff and the distances of each boat from the base of the cliff. 2. **Set up variables and diagram:** Let the height of the cliff be $h$ meters. Let the distance from the foot of the cliff to the boat with angle $30^\circ$ be $x$ meters. Let the distance from the foot of the cliff to the boat with angle $60^\circ$ be $y$ meters. 3. **Use trigonometric relations:** The angles of depression correspond to angles of elevation from the boats to the top of the cliff. For the boat at angle $30^\circ$: $$ \tan 30^\circ = \frac{h}{x} $$ For the boat at angle $60^\circ$: $$ \tan 60^\circ = \frac{h}{y} $$ 4. **Recall values of tangent:** $$ \tan 30^\circ = \frac{1}{\sqrt{3}} $$ $$ \tan 60^\circ = \sqrt{3} $$ 5. **Express $x$ and $y$ in terms of $h$:** $$ x = \frac{h}{\tan 30^\circ} = h \sqrt{3} $$ $$ y = \frac{h}{\tan 60^\circ} = \frac{h}{\sqrt{3}} $$ 6. **Use the total distance between boats:** The boats are on opposite sides, so: $$ x + y = 200 $$ Substitute expressions: $$ h \sqrt{3} + \frac{h}{\sqrt{3}} = 200 $$ 7. **Simplify the equation:** $$ h \sqrt{3} + \frac{h}{\sqrt{3}} = h \left( \sqrt{3} + \frac{1}{\sqrt{3}} \right) = 200 $$ 8. **Combine terms inside parentheses:** $$ \sqrt{3} + \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3}} $$ So: $$ h \cdot \frac{4}{\sqrt{3}} = 200 $$ 9. **Solve for $h$:** $$ h = \frac{200 \sqrt{3}}{4} = 50 \sqrt{3} $$ 10. **Calculate numerical value:** $$ 50 \sqrt{3} \approx 50 \times 1.732 = 86.6 \text{ meters} $$ 11. **Find distances $x$ and $y$:** $$ x = h \sqrt{3} = 50 \sqrt{3} \times \sqrt{3} = 50 \times 3 = 150 \text{ meters} $$ $$ y = \frac{h}{\sqrt{3}} = \frac{50 \sqrt{3}}{\sqrt{3}} = 50 \text{ meters} $$ **Final answers:** - Height of the cliff: $86.6$ meters - Distance of boat at $30^\circ$: $150$ meters - Distance of boat at $60^\circ$: $50$ meters