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Question
Two boats approaching a light house in mid sea from opposite directions observe the angle of elevation of the top of the light house as 30° and 45° respectively. If the distance between the two boats is 150m, find the height of the light house.
Solution
Let the position of the two boats be at points A and C. Let BD be the lighthouse of height h.
Let AD = x. Then, CD = 150 - x
In ΔBAD,
`tan45^circ = "BD"/"AD"`
⇒ `1 = "h"/"X"`
⇒ h = X ...(1)
In ΔBDC,
`tan30^circ = "BD"/"DC"`
⇒ `1/sqrt(3) = "h"/(150 - X)`
⇒ `150 - "X" = sqrt(3)"h"` ....(2)
From (1) and (2),
`150 - "h" = sqrt(3)"h"`
`150 = (sqrt(3) + 1)"h"`
`"h" = 150/(sqrt(3) + 1) xx (sqrt(3) - 1)/(sqrt(3) - 1)`
= `(150(sqrt(3) - 1))/(3 - 1)`
= `75(sqrt(3) - 1)`
= `75 xx 0.732 = 54.9`
Thus, the height of the light house is 54. 9 m.
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