This is the basic and conceptual question from trigonometry
Topic – height and distance
We have given the angle of elevation of top of tower from two point at a distance 5m and 20m from the base of the tower and in the same straight line with it are complementry we have to find the height of tower
Book RS Aggarwal, Class 10, chapter 5C, question no 4.
From figure: Let θ be the angle of elevation.
Given : Angle are complementary, so another angle is (90−θ)o
From right angle ΔCAB:
⇒tanθ=AB/AC
⇒tanθ=h/5
or h=5tanθ...(1)
From another right angle ΔDAB:
⇒tan(90−θ)o=AB/AD
⇒tan(90−θ)o=h/20
or h=20tan(90−θ)o
or h=20cotθ...(2)
Multiply equation (1) and (2), we get
⇒h2=100×tanθ×cotθ
∴h=10 (As tanθ=1/cotθ)
Height of the tower is 10 m.