One of the most important question from trigonometry of height and distance in which we have to find the height of the tower when from the top of a 7m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°
Book RS Aggarwal, Class 10, chapter 14, question no 19
Let the height of tower PQ=hm
the height of building AB=7m
Angle of elevation ∠PAR=60°
Angle of depression ∠RAQ=45° ∠AQB=45°(∵ alternate angle )
Let BQ=AR=xm
In ΔAQB AB/BQ=tan45°
7/x=1
∴x=7m
In ΔPAR, PR/AR=tan60°
(PQ−QR)/x=√3
(h−7)/x=√3
(h−7)/7=√3
∴h=7(√3−1)‘
∴ height of tower PQ=7(√3−1)m
This is the video solution for this question.