From the top of a tower, the angle of depression of the top of a 10 m high building is $60^\circ$. If the distance between the tower and the building is $50\sqrt{3} ext{ m}$, find the height of the tower.

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From the top of a tower, the angle of depression of the top of a 10 m high building is $60^\circ$. If the distance between the tower and the building is $50\sqrt{3} 	ext{ m}$, find the height of the tower.

Accepted Answer

**Solution:**

1. Let the height of the tower be $h$ meters.

2. Given:
   - Height of building = $10 	ext{ m}$
   - Distance between tower and building (base) = $50\sqrt{3} 	ext{ m}$
   - Angle of depression from top of tower to top of building = $60^\circ$

3. The angle of depression from the top of the tower to the top of the building is $60^\circ$. This means the angle of elevation from the building top to the tower top is also $60^\circ$ (alternate interior angles).

4. Consider the right triangle formed by the tower top, the building top, and the horizontal ground distance between them.

5. The vertical height difference between the tower top and building top is $h - 10$.

6. Using the tangent function:
   $$	an 60^\circ = \frac{	ext{Opposite}}{	ext{Adjacent}} = \frac{h - 10}{50\sqrt{3}}$$

7. Since $	an 60^\circ = \sqrt{3}$, we have:
   $$\sqrt{3} = \frac{h - 10}{50\sqrt{3}}$$

8. Multiply both sides by $50\sqrt{3}$:
   $$h - 10 = 50\sqrt{3} 	imes \sqrt{3} = 50 	imes 3 = 150$$

9. Therefore,
   $$h = 150 + 10 = 160 	ext{ m}$$

**Answer:** $160 	ext{ m}$

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