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ABC is a triangular park with AB=AC=100 ...

ABC is a triangular park with AB=AC=100 m .A clock tower is situated at the midpoint of BC. The angle of elevation `alpha " and " beta ` of the top of the tower at A and B, respectively ,are such that cot `alpha` =3.2 cosec `beta` = 2.6 Find the height of the tower .

A

`100/(3sqrt(3))`

B

25

C

20

D

`10sqrt(5)`

Text Solution

Verified by Experts

The correct Answer is:
C
Let h(m) be the height of the tower
Say BM=d

`theta = "cosec"^(-1)2sqrt(2)`
`2sqrt(2) =(sqrt(d)^(2)-d^(2))/h`
`cot phi =sqrt((100)^(2)-d^(2))/h`
`3sqrt(2)=sqrt((100)^(2)-d^(2))/h`
`18h^(2) = 100^(2)-d^(2)`...........(2)

Put the values of `d^(2)` from (1) and (2)
`18h^(2) = 100^(2) -7h^(2) rArr 25h^(2) = 100 xx 100`
h=20 m
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