Question

The angle of elevation of the top of a tower from two point at distance of 4m and 9m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6m.

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Let AB =hm be the tower, AC = 4m and DA =9m.

In triangle ABC we have

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

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

Now in triangle ABD

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

$\tan(90 degree \: - x) = \frac{h}{4}$

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

From equation (I.) and (ii.)

$\frac{h}{ 9} \times \frac{h}{4} = \tan(x) \times \cot(x)$

$\frac{h}{36} 2 = \frac{1}{cot \: x} \times cotx$

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

h² =36

$\sqrt[ + - ]{36}$

$h = \sqrt{36}$

h = 6m