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:
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:
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Use right-angle triangle trigonometry and basic identities like \( \tan(60^\circ) = \sqrt{3} \), \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \).
Updated On: Jun 4, 2025
- 60
- 50
- 40
- 30
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The Correct Option is C
Solution and Explanation
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 = \( h_1 + h_2 = 30 + 10 = 40 \)
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