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The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be 45^{\circ} from a point A on the plane. Let B be the point 30 m vertically above point A.  If the angle of elevation of distance (in m) of the foot of the tower from the point A is:

 

  • Option 1)

    15(3+\sqrt{3})

     

     

     

     

  • Option 2)

    15(5-\sqrt{3})

  • Option 3)

    15(3-\sqrt{3})

  • Option 4)

    15(1+\sqrt{3})

Answers (1)

\tan 30^{\circ}=\frac{K}{d}\Rightarrow d=\sqrt{3}K\cdots \cdots (1)

\tan 45^{\circ}=\frac{K+30}{d}\Rightarrow d=K+30\cdots \cdots (2)

from (1) and (2)

d=\frac{d}{\sqrt{3}}+30=\left ( 1-\frac{1}{\sqrt{3}} \right )d=30

\Rightarrow d=\frac{30\sqrt{3}}{\sqrt{3}-1}\Rightarrow d=\frac{30\sqrt{3}(\sqrt{3}+1)}{2}=15\sqrt{3}(\sqrt{3}+1)

or d=15(3+\sqrt{3})


Option 1)

15(3+\sqrt{3})

 

 

 

 

Option 2)

15(5-\sqrt{3})

Option 3)

15(3-\sqrt{3})

Option 4)

15(1+\sqrt{3})

Posted by

Vakul

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