Question

The height of a temple is 15 metres. 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 \( 5(3 + \sqrt{3}) \) metres.

Hint:

In height and distance problems, always identify the right triangles formed by the line of sight and apply \( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} \).

Answer 1

Step 1: Understanding the figure.
Let \( AB \) be the temple of height \( 15 \, \text{m} \), and \( CD \) be the building on the opposite side of the road. Let the line of sight from \( A \) (top of temple) meet the building at point \( C \) (top of building) and \( D \) (foot of building). Let the distance between the temple and the building (the road width) be \( x \, \text{m} \).
Step 2: From the given data.
\[ \angle CAD = 45^\circ \quad \text{and} \quad \angle CAE = 30^\circ \] where \( E \) is the top of the temple and \( A \) is the point of observation.
Step 3: Use right triangles.
In right triangle \( ABE \): \[ \tan 45^\circ = \frac{BE}{AB} \] \[ 1 = \frac{x}{15} \Rightarrow x = 15 \]
Step 4: For the angle of elevation to top of building.
Let the height of the building be \( h \, \text{m} \). In right triangle \( ACD \): \[ \tan 30^\circ = \frac{h - 15}{x} \] \[ \frac{1}{\sqrt{3}} = \frac{h - 15}{15} \] \[ h - 15 = \frac{15}{\sqrt{3}} = 5\sqrt{3} \] \[ h = 15 + 5\sqrt{3} \] \[ h = 5(3 + \sqrt{3}) \, \text{m} \] Step 5: Final Answer.
\[ \boxed{\text{Height of the building } = 5(3 + \sqrt{3}) \, \text{m}} \]