AP EAMCET · Maths · Heights and Distances
The horizontal distance between a tower and a builidng 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
Answer & Solution
Correct Answer
(C) 40
Step-by-step Solution
Detailed explanation
\(H = 10\sqrt{3} \tan(60^{\circ})\) \(H = 10\sqrt{3} \times \sqrt{3} = 30\) \(h = 10\sqrt{3} \tan(30^{\circ})\) \(h = 10\sqrt{3} \times \frac{1}{\sqrt{3}} = 10\) Sum = \(30 + 10 = 40\)
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