Question

The height of a temple is 15 metre. From the top of the temple, the angle of elevation of the top of a building on the opposite side of the road is \( 30^\circ \) and the angle of depression of the foot of the building is \( 45^\circ \). Prove that the height of the building is \( 5(3 + \sqrt{3}) \) metre.

Hint:

When dealing with angles of elevation and depression, use the tangent function to relate the angles and distances in the right-angled triangles.

Answer 1

Let the height of the temple be \( h_1 = 15 \, \text{m} \), and the height of the building be \( h_2 \). Let the distance between the temple and the building be \( d \). We have two right-angled triangles formed by the temple, the building, and the road. Let's work with these triangles. Triangle 1: Angle of elevation In the first triangle, the angle of elevation from the top of the temple to the top of the building is \( 30^\circ \). Using the tangent function, we can write: \[ \tan(30^\circ) = \frac{h_2 - h_1}{d} = \frac{h_2 - 15}{d}. \] Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have: \[ \frac{1}{\sqrt{3}} = \frac{h_2 - 15}{d} \quad \Rightarrow \quad d = \sqrt{3}(h_2 - 15). \] Triangle 2: Angle of depression In the second triangle, the angle of depression to the foot of the building is \( 45^\circ \). Using the tangent function again, we can write: \[ \tan(45^\circ) = \frac{h_1}{d} = \frac{15}{d}. \] Since \( \tan(45^\circ) = 1 \), we have: \[ 1 = \frac{15}{d} \quad \Rightarrow \quad d = 15. \] Solving for \( h_2 \) Substitute \( d = 15 \) into the equation \( d = \sqrt{3}(h_2 - 15) \): \[ 15 = \sqrt{3}(h_2 - 15). \] Now, solve for \( h_2 \): \[ h_2 - 15 = \frac{15}{\sqrt{3}} = 5\sqrt{3}, \] \[ h_2 = 15 + 5\sqrt{3}. \] Thus, the height of the building is: \[ h_2 = 5(3 + \sqrt{3}) \, \text{metres}. \]
Conclusion: The height of the building is \( 5(3 + \sqrt{3}) \, \text{metres}. \)