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$ABC$ is a triangular park with $AB = AC = 100 m$ . $A$ clock tower is situated at the midpoint of $BC$ . The angles of elevation of the top of the tower at $A$ and $B$ are $\cot^{- 1} 3.2$ and $cose c^{- 1} 2.6$ respectively. Determine the height of the tower.

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16 m

25 m

50 m

none of these

answer is B.

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Detailed Solution

It is given that

AB = AC = 100 m

and the angles of elevation of the top of the tower at

A

and

B

are

\cot^{- 1} 3.2

and

cose c^{- 1} 2.6 .

cotα = 3.2,

cotβ = 2.6

△ ABC is isosceles.
AB = AC, the median AD is perpendicular to base BC.
⇒

DA = hcotα

⇒

DB = hcotβ

⇒

100^{2} = D A^{2} + D B^{2}

⇒

100^{2} = h^{2} co t^{2} α + h^{2} co t^{2} β

⇒

10 0^{2} = h^{2} (co t^{2} α + co t^{2} β)

⇒

100 = h (\sqrt{co t^{2} α + cose c^{2} β - 1}

⇒

100 = h (\sqrt{3 . 2^{2} + 2 . 6^{2} - 1})

⇒

100 = 4 h

∴ h = 25 m

Hence, the correct option is 2.

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