If the angles of elevation of the top of a tower from 2 points at the

1.	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
Answer» 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