Question

The angles of elevation of the top of a tower from two points at distances of 4 m and 9 m from the base of the tower, which are in the same straight line, are complementary. Prove that the height of the tower is 6 m.

Hint:

Height and Distance Problems:Use \( \tan \) and \( \cot \) formulas.

Answer 1

Let the height of the tower be \( h \).
Given angles:
\[ \theta \text{ and } 90^\circ - \theta \] Using tan formula:
\[ \tan \theta = \frac{h}{4}, \quad \tan (90^\circ - \theta) = \cot \theta = \frac{h}{9} \] \[ \frac{h}{4} \times \frac{h}{9} = 1 \] \[ \frac{h^2}{36} = 1 \] \[ h^2 = 36 \] \[ h = 6 \] Thus, the height of the tower is \( \mathbf{6} \) m.