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Two hagstaffs stand on a horizontal plane. A and B are two points on the line joining their feet and between them. The angles of elevation of the tops of the flagstaffs as seen from A are `30^@` and` 60^@` and as seen from B are `60^@`and `45^@`. If AB is 30 m, then the distance between the flagstaffs is

A

`30+15sqrt3`

B

`45+15sqrt3`

C

`60-15sqrt3`

D

`60+15sqrt3`

Text Solution

Verified by Experts

The correct Answer is:
D

Let x and y be the hight of the flagstaffs at P and Q respectively.

Then `AP= x cot 60 ^@ = (x)/(sqrt(3))`
`AQ = y cot 30 ^@= y sqrt(3)`
`BP= x cot 45 ^@ = x`
and `BQ= y cot 60 ^@ = (y)/(sqrt(3)`
`therefore AB=BP- AP= x -(x)/(sqrt(3)`
`rArr 30 sqrt(3)=(sqrt(3)-1)x`
`rArr x= 15 (3 + sqrt(3))`
`Similarly, 30 =y(sqrt(3)-1/sqrt(3))`or `y= 15 sqrt(3)`
`therefore PQ= BP+BQ= x+ y/(sqrt(3))`
`= 15(3+ sqrt(3))+15`
`= (60 + 15 sqrt(3))m`
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