Question

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. \( \frac{1}{2} b \tan\alpha \sec B \)
• B. \( \frac{1}{2} b \tan\alpha \csc A \)
• C. \( \frac{1}{2} c \tan\alpha \sin C \)
• D. \( \frac{1}{2} a \tan\alpha \csc A \)

Hint:

If the angle of elevation to a point is the same from three vertices of a horizontal triangle, the foot of the perpendicular from that point is the circumcenter of the triangle. The distance from the foot to any vertex is the circumradius (R). Use the formula \( h = R\tan\alpha \) and the sine rule \( 2R = a/\sin A \) to connect the height to the triangle's properties.

Answer 1

The Correct Option is D

Let the top of the hill be P and its foot on the horizontal plane be O. Let the height of the hill be \( h = PO \). The angle of elevation from A, B, and C to P is \( \alpha \). This means \( \angle PAO = \angle PBO = \angle PCO = \alpha \). In the right-angled triangles \( \triangle PAO, \triangle PBO, \triangle PCO \), we have: \[ OA = \frac{h}{\tan\alpha} = h\cot\alpha \] \[ OB = h\cot\alpha \] \[ OC = h\cot\alpha \] Since \( OA = OB = OC \), the point O (the foot of the hill) is equidistant from the vertices A, B, and C of the horizontal triangle. This means O is the circumcenter of \( \triangle ABC \). The distance from the circumcenter to any vertex is the circumradius, R. So, \( R = h\cot\alpha \), which implies \( h = R\tan\alpha \). Now we need to express R in terms of the sides and angles of \( \triangle ABC \). We use the sine rule, which relates the sides to the circumradius: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \] From this, we can express R in several ways, for example, \( R = \frac{a}{2\sin A} \). Substitute this expression for R back into the equation for height: \[ h = \left(\frac{a}{2\sin A}\right) \tan\alpha = \frac{1}{2} a \tan\alpha \frac{1}{\sin A} \] \[ h = \frac{1}{2} a \tan\alpha \csc A \] This matches option (D).