The angle of elevation of the top of a building at a distance of $70 ext{ m}$ from its foot on a horizontal plane is found to be $60^\circ$. Find the height of the building.

Question

The angle of elevation of the top of a building at a distance of $70 	ext{ m}$ from its foot on a horizontal plane is found to be $60^\circ$. Find the height of the building.

Accepted Answer

**Solution:**

1. Model the situation as a right-angled triangle where the height of the building $h$ is the side opposite to the angle of elevation $60^\circ$, and the distance from the foot of the building $70 	ext{ m}$ is the adjacent side.

2. Use the tangent function:
   $$	an(	heta) = \frac{	ext{opposite}}{	ext{adjacent}} = \frac{h}{70}$$

3. Substitute $	heta = 60^\circ$ and $	an(60^\circ) = \sqrt{3}$:
   $$\sqrt{3} = \frac{h}{70}$$

4. Solve for $h$:
   $$h = 70 	imes \sqrt{3}$$

**Answer:** The height of the building is $70\sqrt{3} 	ext{ m}$.

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