Question

The shadow of a tower , when the angle of elevation of a sun is 45° , is found to be 10 metres longer than when the angle of elevation is 60°. find the height of the tower. [ take √3 = 1.732 ]

Answer 1

Let AB be the tower and let AC and AD be it's shadows when the angle of elevation of the sun 60° and 45° respectively.


Therefore,

/_ ACB = 60° , /_ ADB = 45° , /_ DAB = 90° and CD = 10 m.



Let AB = h metres and AC = x metres.

From right ∆ CAB , we have


AB / AD = tan60° = √3



h/ x = √3



X = h/ √3 ---------(1)


From right ∆ DAB , we have:


AB / AD = tan 45° = 1





=> h / 10 + x = 1




=> 10 + x = h



=> x = ( h -10 ) ------(2)


From (1) and (2), we get :


h/ √3 = h -10


=> h ( √3 - 1 ) = 10√3



=> h = 10√3 / ( √3 - 1 )


=> h = 10√3 / ( √3 - 1 ) × ( √3 + 1 )/(√3 + 1 )



=> h = ( 15 + 5√3 )



=> h = ( 15 + 5 × 1.732 )


=> h = 23.66 m.



Hence,



The height of the tower is 23.66 m.

Answer 2

Hey mate ^_^

=======
Answer:
=======

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 triangle DAB,

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

#Be Brainly❤️