Question

The angle of elevation of the top of the building from a point 10 meters away from the base of the building is 60°, then the height of the building is:

• A. \( \frac{10}{\sqrt{3}} \, \text{m} \)
• B. \( 10\sqrt{3} \, \text{m} \)
• C. \( 10 \, \text{m} \)
• D. \( \frac{10}{3} \, \text{m} \)

Hint:

For right-angled triangles, use the tangent function to relate the angle of elevation or depression with the height and distance.

Answer 1

The Correct Option is B

We are given that the distance from the point to the base of the building is 10 meters, and the angle of elevation to the top of the building is 60°. We need to find the height of the building. Step 1: Represent the situation in a right triangle. - The height of the building is the opposite side. - The distance from the point to the base of the building is the adjacent side. - The angle of elevation is \( 60^\circ \). Step 2: Use the tangent of the angle of elevation to find the height \(h\): \[ \tan 60^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{10} \] Step 3: We know that \( \tan 60^\circ = \sqrt{3} \), so: \[ \sqrt{3} = \frac{h}{10} \] Step 4: Solve for \(h\): \[ h = 10\sqrt{3} \] Thus, the height of the building is \( 10\sqrt{3} \, \text{m} \).