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 is 60°. find the height of 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

Answer 2

Answer:

Let length of shadow when it is 60 be x

Then x+10=shadow when 45

Tan60 =h/x

Root3x = h

Then root3x=10+x

Root 3x-x=10

X=10[root3+1]/2

H=5(root 3+1)× root3

H=23.66

Step-by-step explanation: