Question

PQ is a vertical tower. P is the foot, Q the top of the tower, A, B, C are three points in the horizontal plane through P. The angles of elevation of Q from A, B, C are equal and each is equal to $\theta$. The sides of the triangle ABC are a, b, c and the area of the triangle ABC is $\Delta$. The height of the tower is

• A. (abc) tan $\theta /4\Delta$
• B. (abc) sin $\theta /4\Delta$
• C. (abc) cot $\theta /4\Delta$
• D. None

Answer 1

The correct option is A (abc) tan $\theta /4\Delta$

Let the height PQ of the tower be h. Since the angles of elevation of Q from each of the points A, B, C is $\theta$, we have
PA = PB = PC = h cot $\theta$ ...(1)
Since P is equidistant from A, B and C it is the circumcentre of the $\Delta$ ABC.
PA = PB = PC = circumradius of $\Delta$ ABC = abc/4$\Delta$ ...(2)
Hence from (1) and (2),
h = abc tan $\theta$/4$\Delta$