Question

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

Answer 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