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Question
An observer point for ships moving in the sea 500m above the sea level. The person manning this point observes the angle of depression of twc boats as 45° and 30°. Find the distance between the boats when they are on the same side of the observation point and when they are on opposite sides of the observation point.
Solution
Case 1: When the boats are on same side of the observation point.
Let the position of the two ships be C and D. Let A be the point of observation.
AB = 500 m
In ΔBAC,
`tan45^circ = "AB"/"BC"`
⇒ `1 = 500/"BC"`
⇒ BC = 500 ....(1)
In ΔABD,
`tan30^circ = "AB"/"BD"`
⇒ `1/sqrt(3) = 500/"BD"`
⇒ `"BD" = 500sqrt(3)` ...(2)
From (1) and (2),
`"CD" = "BD" - "BC" = 500(sqrt(3) - 1) = 500 xx 0.732 = 366`
Thus, in this case, the distance between the boats is 366 m.
Case 2: When the boats are on different side of the observation point.
Let the position of the two ships be A and C. Let B be the point of observation.
In ΔBAD,
`tan45^circ = "BD"/"AD"`
⇒ `1 = 500/"AD"`
⇒ AD = 500 ....(1)
In ΔBDC,
`tan30^circ = "BD"/"DC"`
⇒ `1/sqrt(3) = 500/"DC"`
⇒ `"DC" = 500sqrt(3)` ....(2)
From (1) and (2),
`"AC" = "AD" + "DC" = 500 (1 + sqrt(3)) = 500 xx 2.732 = 1366`
Thus, in this case, the distance between the boats is 1366 m.
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