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Two man are on the opposite sides of a tower. They measure the angles of elevation the top of the tower as `30^(@)" and "60^(@)`. If the height of the tower is 150 m, find the distance between the two men.

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Let CD be a tower of height 150 m. Two men A and B are on the opposite sides of the tower. Given that
`angleDAC=60^(@)" and "angle=30^(@)`
In `DeltaDAC`
`tan60^(@)=(DC)/(AC)rArrsqrt3=150/(AC)`
`rArr AC=150/sqrt3=50sqrt3 m`
In `DeltaBCD`
`tan30^(@)=(DC)/(BC)rArr1/sqrt3=150/(BC)rArrBC=150sqrt3 m`
`:. AB=AC=BC` ltBRgt `=(50sqrt3+150sqrt3)m=200sqrt3m`
Therefore the distance between two men `=200sqrt3 m`
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