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Question
From the top of a tower 100 m high a man observes the angles of depression of two ships A and B, on opposite sides of the tower as 45° and 38° respectively. If the foot of the tower and the ships are in the same horizontal line find the distance between the two ships A and B to the nearest metre.
(Use Mathematical Tabels for this question)
Solution
Let AC = x,
BC = y
Now ∠A = 45° ...(Alternative angles)
and ∠B = 38° ...(Alternative angles)
In ΔACD,
tan 45° = `100/x`
`\implies` 1 = `100/x`
∴ x = 100 ...(1)
In ΔBCD,
tan 38° = `100/y`
0.7812 = `100/y`
y = `100/0.7812 ≈ 128`
Hence Distance between two ships A, B
= x + y
= 100 + 128
= 228 m
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