Question

The angle of elevation of top of the tower from two points P and Q at distance of a and b respectively from the base and in same straight line with it are complementary. Prove that the height of tower is √ab?

Answer 1

ANSWER....

Let PQ be the tower and let R and S be the two position of the observer.

PR = a meters
QS = b meters

Let angle PRQ = theta
angle PSQ = 90° - theta

Let PQ = h meters
from right triangle RPQ,

$\frac{pq}{pr} = \tan(theta)$
$\frac{h}{a} = \tan(theta)$
h = a tan theta

from right triangle DAB

$\frac{pq}{ps} = \tan(90 - theta)$
$= > \frac{h}{b} = \cot(theta)$

from (i) and (ii),
${h}^{2} = ab$
$h = \sqrt{ab}$

the height of tower =
$\sqrt{ab} \: meters$



$\red{hope \: it \: helps}$
$\blue{thanks}$

Answer 2

Answer:

Check your answer please