Question

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

Answer 1

Let angle ACB be alpha and angle ADB be beta

As they are complimentary so beta =90-alpha

Now tan alpha =height /base

=Ab/4

Tan beta =AB/9

Tan 90-alpha=ab/9

Cot alpha=ab/9

1/tan alpha=ab/9

4/AB=ab/8

Ab square=36

Ab=6

Answer 2

let AB = hm be the tower,

AC= 4m and DA = 9m

In ∆ABC we have

$\frac{p}{b} = \frac{ab}{ac}$

$tan \: x \: = \frac{h}{9}$......(I.)

now in ∆ABD

$\frac{p}{b} = \frac{ab}{da}$

$\tan(90° - x) = \frac{h}{4}$

$\cot(x) = \frac{h}{4}$....(ii.)

from equation i &ii

$\frac{h}{9} \times \frac{h}{4} = \frac{1}{cot \: x} \times cot \: x$

$\frac{ {h}^{2} }{36} = 1$

h² =36

h=±√36

h=√36

h=6m

hence the height of tower is 6m.