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Question
From the top of a tower h m high, the angles of depression of two objects, which are in line with the foot of the tower are α and β (β > α). Find the distance between the two objects.
Solution
Given: the height of tower is h m.
∠ABD = α and ∠ACD = β
Let CD = y and BC = x
In ∆ABD,
tan α = `"AD"/"BD"`
⇒ tan α = `"h"/("BC" + "CD")`
⇒ tan α = `"h"/(x + y)`
⇒ x + y = `"h"/tan α`
⇒ y = `"h"/tan α - x` ...[Equation 1]
In ∆ACD,
tan β = `"AD"/"CD"`
⇒ tan β = `"h"/y`
⇒ y = `"h"/tan β` ...[Equation 2]
Comparing equation 1 and equation 2,
`"h"/tan α - x = "h"/tan β`
⇒ x = `"h"/tan α - "h"/tan β`
⇒ x = `"h"(1/tan α - 1/tan β)`
⇒ x = h(cot α – cot β)
Hence, we have got the required distance between the two points, i.e. h(cot α – cot β)
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