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Question
From a point, 36 m above the surface of a lake, the angle of elevation of a bird is observed to be 30° and the angle of depression of its image in the water of the lake is observed to be 60°. Find the actual height of the bird above the surface of the lake.
Solution
Let A be a point 36 m above the surface of the lake and B be the position of the bird. Let B' be the image if the brid in the water.
Here, AC = DE = 36 m, ∠BAE = 30° and ∠B’AE = 60°,
Let BE = h m,
Then, B'D = BD = 36 + h ...(∴ B’ is image of B about D)
∴ B’E = B’D + DE
= 36 + 36 + h
= 72 + h ...(i)
In ΔABE,
`(BE)/(AE) = tan 30^circ`
`=> AE = sqrt(3)h` ...(ii)
In ΔAB’E,
`(B^’E)/(AE) = tan 60^circ`
`=> (72 + h )/(AE) = sqrt(3)` ...[From (i)]
`=> 72 + h = (sqrt(3)h)sqrt(3)` ...[From (ii)]
`=>` 72 + h = 3h
∴ h = 36 m
Hence, the actual height of the bird above the surface of the lake = 36 + 36 = 72 m.
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