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The shadow of tower when the angle of elevation of sun is 45° is found to be 10 m longer than when it is 60°. find the height of tower ?​

Answers (1)

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

Posted by

Satyajeet Kumar

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