Question : The length of the shadow of a vertical tower on level ground increases by 10 m when the altitude of the sun changes from 45° to 30°. The height of the tower is:
Option 1: $10 \sqrt{3}$ m
Option 2: $5 \sqrt{3}$ m
Option 3: $5(\sqrt{3}+1)$ m
Option 4: $10(\sqrt{3}+1)$ m
Correct Answer: $5(\sqrt{3}+1)$ m
 
  Solution :
 
 
 
  Let the height of the tower be $h$ meters.
  
  And $p$ is the shadow of the tower
  
  In $\triangle ABC$,
  
  ⇒ $\tan45^\circ = \frac{h}{p}$
  
  ⇒ $p = h$....................................(equation i)
  
  In $\triangle ABD$
  
  ⇒ $\tan30^\circ = \frac{h}{10+p}$
  
  Putting the value $p$ from equation (i), we get:
  
  ⇒ $\frac{1}{\sqrt3} = \frac{h}{h+10}$
  
  ⇒ $10 + h = \sqrt3h$
  
  ⇒ $h(\sqrt3 - 1) = 10$
  
  ⇒ $h = \frac{10}{(\sqrt3 - 1)}$
  
  On rationalisation
  
  ⇒ $h = \frac{10(\sqrt3 + 1)}{(\sqrt3 - 1)(\sqrt3 + 1)}$
  
  ⇒ $h = \frac{10(\sqrt3 + 1)}{2}$
  
  ⇒ h= $5(\sqrt{3}+1)$ m
  
  Hence, the correct answer is $5(\sqrt{3}+1)$ m.
 
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