Question

The horizontal distance between a tower and a building is \( 10\sqrt{3} \) units. If the angle of depression of the foot of the building from the top of the tower is \( 60^\circ \) and the angle of elevation of the top of the building from the foot of the tower is \( 30^\circ \), then the sum of the heights of the tower and the building is:

• A. 60
• B. 50
• C. 40
• D. 30

Hint:

Use right-angle triangle trigonometry and basic identities like \( \tan(60^\circ) = \sqrt{3} \), \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \).

Answer 1

The Correct Option is C

Let the distance between the tower and the building be \( AB = 10\sqrt{3} \). Let \( C \) be the top of the tower, and \( D \) be the top of the building.
From triangle \( \triangle CAB \): \[ \tan(60^\circ) = \frac{h_1}{10\sqrt{3}} \Rightarrow \sqrt{3} = \frac{h_1}{10\sqrt{3}} \Rightarrow h_1 = 30 \] From triangle \( \triangle DAB \): \[ \tan(30^\circ) = \frac{h_2}{10\sqrt{3}} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h_2}{10\sqrt{3}} \Rightarrow h_2 = 10 \] So, the total height = \( h_1 + h_2 = 30 + 10 = 40 \)