Question

if the angles of elevation of the top of a tower from 2 points at the distance of 4m and 25m from the base of the tower and in the same line with it are complimentary, what is the hight of tower
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Answer 1

Step-by-step explanation:

solution:-

given AB is tower

p and q are the point of 4m and 25 m

from fig PB=4m ,BQ=25m

$angle \: of \: elevation \: from \: p \: be \: \alpha$

$and \: angle \: of \: elevation \: from \: q \: be \: \beta$

$given : \alpha + \beta = 90 \degree$

in triangle ABP

$\tan( \alpha ) = \frac{ab}{bp}$

in triangle ABQ

$\tan( \beta ) = \frac{ab}{bq}$

now we write as

$\beta = 90 - \alpha$

so put the value on tan(beta), we get

$\tan(90 - \alpha ) = \frac{ab}{bq}$

now , we get

$\cot( \alpha ) = \frac{ab}{bq}$

$\frac{1}{ \tan( \alpha ) } = \frac{ab}{bq}$

$\tan( \alpha ) = \frac{bq}{ab}$

now

$\tan( \alpha ) = \frac{ab}{bp} = \tan( \alpha ) = \frac{bq}{ab}$

so we can write :-

$\frac{ab}{bp} = \frac{bq}{ab}$

use cross multiplication

AB²=BQ×BP

AB²=25×4

AB²=100

AB=10m

height of tower is 10 m