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Question
Two ships are sailing in the sea on either side of the lighthouse. The angles of depression of two ships as observed from the top of the lighthouse are 60° and 45° respectively. If the distance between the ships is `200[(sqrt(3) + 1)/sqrt(3)]` metres, find the height of the lighthouse.
Solution
Let A and B the position of the first ship and the second ship
Distance = `200((sqrt(3) + 1)/sqrt(3))`m
Let the height of the lighthouse CD be h
In the right ∆ACD, tan 60° = `"CD"/"AD"`
`sqrt(3) = "h"/"AD"`
∴ AD = `"h"/sqrt(3` ...(1)
In the right ∆BCD
tan 45° = `"DC"/"BD"`
1 = `"h"/"BD"`
∴ BD = h
Distance between the two ships = AD + BD
`200((sqrt(3) + 1)/sqrt3) = "h"/sqrt3 + "h"`
⇒ `200 (sqrt(3) + 1) = "h" + sqrt(3)"h"`
`200(sqrt(3) + 1) = "h"(1 + sqrt(3))`
⇒ h = `(200(sqrt(3) + 1))/((1 + sqrt(3))`
h = 200
Height of the light house = 200 m
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