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From the top of a lighthouse, an observ...

From the top of a lighthouse, an observer looking at a boat makes an angle of depression of `60^(@)`. If the height of the lighthouse is `90m`, then find how far is the boat from the lighthouse.
`(sqrt(3)=1.73)`

Text Solution

Verified by Experts

Let seg AB represent the lighthouse.
C is the position of the ship.
AD is the horizontal line and `angleDAC` is the angle of depression.
AB = 90 m and `angleDAC = 60^@` ….(Given)
`angleACB and angleDAC` are alternate angles.
`therefore angleACB = angleDAC = 60^@`.
In right angled `triangleABC`,
`tan 60^@ = (AB)/(BC)` ....(By definition )
`therefore sqrt(3) = 90/(BC)` ...(`tan 60^@ = sqrt(3)`)
`therefore BC = 90/sqrt3`
`therefore BC = 90/sqrt3 xx sqrt3/sqrt3`
`therefore BC = (90sqrt(3))/(3)`
`therefore BC = 30sqrt(3)`
`therefore BC = 30 xx 1.73` (`therefore " "sqrt3 = 1.73)`
`therefore BC = 51.9 m
The ship is at the distance of 51.9 m from the lighthouse.
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