Question

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

Answer 1

Step-by-step explanation:

multiply 1and 2

h^2=. 36. (cot theta and tan theta are cancelled)

hence,. h=6 m. (taking square root)

mark me BRAINLIEST

Answer 2

Given :-

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

To Prove :-

Prove that the height of the tower is 6 m.

Solution :-

(Refer to the attachment for the figure)

In right ΔABC,

$\sf tan \ x = \dfrac{AB}{BC}$

$\sf tan \ x = \dfrac{AB}{4}$

$\sf AB = 4 \ tan \ x \qquad ...(1)$

Again, from right ΔABD,

$\sf tan \ (90^{o} -x) =\dfrac{AB}{BD}$

$\sf cot \ x =\dfrac{AB}{9}$

$\sf AB = 9 \ cot \ x \qquad ...(2)$

Multiplying equation (1) and (2), we get

$\sf AB^{2} = 9 \ cot \ x \times 4 \ tan \ x$

$\implies \sf AB^{2}=36 \ ( \because cot \ x=\dfrac{1}{tan \ x} )$

$\sf \implies AB= \pm \ 6$

Since height cannot be negative. Therefore, the height of the tower is 6 m.

Hence Proved!