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.

Answer 1

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)