Question 49

Mathematics Heights and Distances Hard

If the angle of elevation of the top of a hill from each of the vertices A, B and C of a horizontal triangle is \(\alpha\), then the height of the hill is

(A) \(\frac{1}{2}b\, \tan \alpha\, secB\)
(B) \(\frac{1}{2}b\, \tan \alpha\, cosecA\)
(C) \(\frac{1}{2}c\, \tan \alpha\, \sin C\)
(D) \(\frac{1}{2}a\, \tan \alpha\, cosecA\)
✓ Correct Answer: Option D Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

Since the angle of elevation from the three vertices is the same, denoted by \( \alpha \), the tower must be equidistant from the three vertices, meaning the tower is at the circumcenter of the triangle. The distance of the circumcenter from any vertex is: \[ h \cot \alpha = R \quad \text{(circum-radius)}. \] We know the following relationship in a triangle: \[ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = \frac{1}{2R}. \] Hence, \[ h \cot \alpha = \frac{a}{2 \sin A} \quad \Rightarrow \quad h = \frac{a}{2} \tan \alpha \csc A. \]

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