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

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