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Question
As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Solution
Let AB be the lighthouse and the two ships be at point C and D respectively.
In ΔABC,
`"AB"/"BC"` = tan 45°
`75/"BC"` = 1
BC = 75 m
In ΔABD,
`"AB"/"BD"` = tan 30°
`75/("BC" + "CD") = 1/sqrt3`
`75/(75+"CD") = 1/sqrt3`
`75sqrt3 = 75 + "CD"`
`75(sqrt3 -1)"m" = "CD"`
If we take the value of `sqrt3` = 1.73
`75(1.73 -1)"m"`
= 54.91 m.
Hence, the Distance between the two ships is 54.91 m.
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