COMEDK · Maths · 18. Heights and Distances
The angle of elevation of the top of a TV tower from three points \(A, B\) and \(C\) in a straight line through the foot of the tower are \(\alpha, 2 \alpha\) and \(3 \alpha\) respectively. If \(A B=a\), then height of the tower is
- A \(a \tan \alpha\)
- B \(a \sin \alpha\)
- C \(a \sin 2 \alpha\)
- D \(a \sin 3 \alpha\)
Answer & Solution
Correct Answer
(C) \(a \sin 2 \alpha\)
Step-by-step Solution
Detailed explanation
Let \(D E\) be the height of the tower. Now, in \(\triangle A B E\), \[ \begin{aligned} \angle A E B &=\angle E B C-\angle E A B \\ &=2 \alpha-\alpha=\alpha \\ \therefore \quad \angle A E B &=\angle B A E \\ \Rightarrow \quad B E &=A B=a \end{aligned} \] Now,…
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