The angles of elevation of the top of a tower from two points on the ground at distances 32 m and 18 m from its base and in the same straight line with it are complementary. The height(in m) of the tower is .....

Let $$ \angle ADB = \theta $$ 

then $$\angle ACD = (90^\circ - \theta) $$

In right triangle $$ \triangle ABD $$

$$ \tan \theta = \frac{AB}{BD} $$

$$\tan \theta = \frac{AB} {32} $$...........(1)

$$\triangle ABC $$ 

$$ \tan(90^\circ - \theta) = \frac{AB}{BC} $$

$$ \cot \theta = \frac{AB}{18} $$ ...........(2)

Multiplying Equation (1) and Equation (2)

$$ \tan \theta \times \frac{1}{\tan \theta} = \frac{AB}{32} \times \frac{AB}{18} $$

$$ 1 = \frac{(AB)^2} {32 \times 18} $$

$$ AB = \sqrt {18 \times 32 } $$

 $$ \Rightarrow AB = 2 \times 2 \times 2 \times 3 $$

$$ \Rightarrow AB = 24 $$ 

The height of tower = 24 m Ans