Question

23. The angles of elevation of the top of a tower from two points at a distance of 'a' m and 'b' 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 (ab)^1/2

Answer 1

HEY

HERE IS ANSWER.

Question:

The angles of elevation of the top of a tower from two points at a distance of 'a' m and 'b' 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 (ab)^1/2.

Given:

The angles of elevation of the top of a tower from two points at a distance of 'a' m and 'b' 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 (ab)^1/2.

Proof.
A
|\ \
| \ \
| \ \
| \ \
| \ \
| \ \
|_____\ \
B.... a .... C D
|____ b___|



$let /angle{C}be\theta$

hence

$\angle {D}=90°-\theta$

let height be m

hence.

$tan \theta = \frac{m}{a}.......(1)$

$tan \: 90\degree- \theta =tan90 \degree - \theta= \frac{m}{b} .......(2)$


multiply equation 1 and 2

$tan \theta \times sec\theta = \frac{m}{b} \times \frac{m}{a}$

1= m^2/ab

m^2=ab


hence.

$m = \sqrt{ab}$

hence

$\bf{m = {ab}^{ \frac{1}{2} }}$



Hope it helps

thanks

Answer 2

Answer:

Check your answer please