sir this is the question from the book -ML aggarwal( avichal publication) class 10th , chapter20 , heights and distances. IT’s very important question we have been given that An observer 1.5 m tall is 20.5 metres away from a tower 22 metres high.
we are asked to Determine the angle of elevation of the top of the tower from the eye of the observer.
question no 13 , heights and distances , ICSE board
In the figure,
AB is the tower and CD is an observer
θ is the angle of observation
It is given that
AB = 22m
CD = 1.5 m
Distance BD = 20.5 m
From the point C construct CE parallel tp DB
AE = 22 – 1.5 = 20.5 m
CE = DB = 20.5 m
tan θ = AE/CE
Substituting the values
tan θ = 20.5/20.5 = 1
tan 450 = 1
θ = 450