Question

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

Answer 1

the shadow of a tower when the angle of elevation of sun is 45 degrees is found to be 10m longer than when it was 60degrees.find height of tower
Answer:-
in:- (image)

Answer 2

$\large{\underline{\bf{\pink{Answer:-}}}}$  

  ✰ height of the tower is 23.66m

$\large{\underline{\bf{\blue{Explanation:-}}}}$  

  $\large{\underline{\bf{\green{Given:-}}}}$

✰ Angle of elevation of sun = 45°

✰ shadow is found 10m when the angle of elevation of sun is 60°

$\large{\underline{\bf{\green{To\:Find:-}}}}$

✰ we need to find the height of the tower.

$\huge{\underline{\bf{\red{Solution:-}}}}$  

  Let AB be the tower Let AC and AD be the shadows when the angles of elevation of sun are 60° and 45°

$\angle\:ACB=60\degree\:,\:\angle\:ADB=45\degree,$

$\angle\:DAB=90\degree\:,\:and\:CD=10m$

Let AB = h meters and AC = x meters.In right angled triangle∆ CAB,

$:\implies\:\frac{AC}{AB}=cot 60\degree$

$:\implies\:\pink{cot 60\degree= \frac{1}{\sqrt{3}}}$

$:\implies\:\frac{x}{h}=\frac{1}{\sqrt{3}}$

$:\implies\:x=\frac{h}{\sqrt{3}}........(1$

In right angled triangle∆DAB

$\frac{AD}{AB}=cot 45\degree$

$:\implies\:\pink{cot 45\degree= {1}}$

$:\implies\:\frac{x+10}{h}=1$

$:\implies\:x+10=h$$:\implies\:x=h- 10.........(2)$

Substituting the value of x from (i) in (ii)

$:\implies\:\frac{h}{\sqrt{3}}=h-10$

$:\implies\:h=\sqrt{3}h-10\sqrt{3}$

$:\implies\:(\sqrt{3}-1)h=10\sqrt{3}$

$:\implies\:\frac{10\sqrt{3}}{(\sqrt{3}-1)}$

Rationalising the denominator:-

$:\implies\:h=[\frac{10\sqrt{3}}{(\sqrt{3}-1)}\times\frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}]$

$:\implies\:h= 5\sqrt{3}(\sqrt{3}+1)$

$:\implies\:h=15+5\sqrt{3}$

$:\implies\:h=(5+5\times1.732)=(15+8.66)$

$:\implies\:\bf\purple{h= 23.66}$

Hence , the height of the tower is 23.66m.

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