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The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Find the height of the tower. 

Answers (1)

Solution.         According to question
                      
     Let the height of tower = h
  \tan 60^{\circ}= \frac{h}{BD}          \left [ \because \tan \theta = \frac{Perpendicular}{Base} \right ]
\sqrt{3}= \frac{h}{BD}                      \left [ \because \tan 60^{\circ}= \sqrt{3} \right ]
BD= \frac{h}{\sqrt{3}}\cdots \left ( 1 \right )
\tan 30^{\circ}= \frac{h}{BC}= \frac{h}{BD+DC}       \left [ \because BC= BD+DC \right ]
\frac{1}{\sqrt{3}}= \frac{h}{BD+50}
BD+50= \sqrt{3h}    

[by cross multiplication]
\frac{h}{\sqrt{3}}+50= \sqrt{3h}

  [from equation (1)]
\frac{h}{\sqrt{3}}= \sqrt{3h}-50
h= 3h-50\sqrt{3}
h= \frac{50\sqrt{3}}{2}
h= 25\sqrt{3}m
 

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infoexpert27

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