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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)

best_answer

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

Posted by

Aadil

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