The shadow of a tower, when the angle of elevation is 54 is found to be 15 meters longer than when it is 70 Find the height of the tower.
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Let AB be the tower with height h.
Let AC and AD be the shadows when elevation of sun are 60 degrees and 45 degrees.
As per given, CD=10m
let us assume CA=x
In triangle ACB,
tan60°=opposite side /adjacent side
√3=h/AC
√3=h/x
x=h/√3 ------ equation (1)
In traingleDAB,
tan45°=AB/AD
=h/(AC+DC)
1=h/(x+10)
x+10=h-----equation(2)
By substituting the value of x in equation 2 we get:
h/√3+10=h
h-h/√3=10
h√3-h=10√3
h(√3-1)=10√3
h=10√3/√3-1
Rationalizing factor is √3+1
h=10√3(√3+1)/[(√3-1)x(√3+1)]
h=10√3(√3+1)/(3-1)
h=10√3(√3+1)/2
h=5√3(√3+1) m
h=5(3+√3)
=15+5*√3
=15+5*1.732
=15+8.660
=23.66 m
∴ Height of tower is 23.66m
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