ABC is a triangular park with AB = AC = 100 metres.A vertical tower is situated at the midpoint of BC. if the angles of elevation of the top of the tower at A and B are  \cot^{-1}(2\sqrt2) respectively , then the height of the tower (in meters) is : 

  • Option 1)

    \frac{100}{3\sqrt3}

  • Option 2)

    10\sqrt5

  • Option 3)

    20

  • Option 4)

    25

Answers (1)

 

AM = y ; MD = h ( tower height )

y^{2}=100-x^{2}............................(1)

cot\theta=3\sqrt2  =>  tan\theta=\frac{1}{3\sqrt2}

=> \frac{h}{y}=\frac{1}{3\sqrt2}

=> {y}={3\sqrt2}h

from (1)

18h^{2}=100-x^{2}.............................(2)

tan\phi =\frac{1}{\sqrt7}=\frac{h}{x}

h{\sqrt7}={x}..........................................(3)

100^{2}-7h^{2}=18h^{2}

100^{2}=25h^{2}

\frac{100}{5}=h

20=h

correct option(3)

 

 


Option 1)

\frac{100}{3\sqrt3}

Option 2)

10\sqrt5

Option 3)

20

Option 4)

25

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