The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m, find the height of the tower from the base of the tower and in the same straight line with it are complementary.

Step 1: Analyse the give data and construct an imaginary figure by the given hints.
GIVEN: The angles of elevation of the top of the tower from the two points
are complementary angles.
Let, if you say angle of elevation \angle ADC = \theta
then, angle of elevation \angle ACB = 90\degree - \theta
The distance between the pole and points
BC = 4 cm
DB = 9m
Height of the tower AB = ?
Step 2: Determine the height of the tower by using the trigonometric ratios
From the figure:
Take right angle triangle ADB
\tan \theta = \frac{opposite}{adjacent} = \frac{AB}{DB}
\tan \theta = \frac{AB}{9}.....................(1)
Take right triangle ACD
\tan (90\degree - \theta) = \frac{AB}{CB}
\cot \theta = \frac{AB}{4}.................(2) (\because \tan (90\degree - \theta) = \cot \theta )
Step 3: Multiplying equation (1) and (2)
\tan \theta = \frac{AB}{9}..................(1)
\cot \theta = \frac{AB}{4}.................(2)
Multiply
\cot \theta * \tan \theta = \frac{AB}{4} * \frac{AB}{9}
\frac{1}{\tan \theta} *\tan \theta = \frac{AB^2}{36} (\because \cot \theta = \frac{1}{\tan \theta} )
1 = \frac{AB^2}{36}
AB^2 = 36
AB = \sqrt{6*6}
AB = 6 m
The height of the tower = 6 m