Q. From the top of a cliff 25 metres high, the angle of elevation of the top of a tower found to be equal to the angle of depression of the foot of the tower, then the height of the tower is.
Solution:
AB is the cliff of height 25 m and PQ is the tower.
Let $\Delta$PBL = $\Delta$QBI. = $\theta$
From rt. $\Delta d \, \Delta$QAB, $\frac{25}{x}$ = tan $\theta \, \therefore \, x$ = 25 cot $\theta$
From $rt \, \Delta d \, \Delta $BPL, $\frac{PL}{x}$ = tan $\theta$
$\therefore $ PL = $x$ tan $\theta$ = 25.cot $\theta$ tan $\theta$ = 25
$\therefore $ Height of the tower -Q P = QL + I.P
= AB + PL = 25 + 25 = 50 m
