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# A man notices two objects in a straight line due west. After walking a distance c due north he observes that the objects subtend an angle α at his eye; and , after walking a further distance 2c due north, an angle β. Show that the distance between the objects is 8c/3 cot β – cot α ; the height of the man is being ignored.

Jitender Pal
7 years ago
Hello Student,
Let the man initially be standing at ‘A’ and ‘B’ be the position after walking a distance ‘c’, so total distance becomes 2c and the objects being observed are at ‘C’ and ‘D’.
Now we have OA = c, AB = 2c
Let CO = x and CD = d
Let ∠CAD = α and ∠CBD = β
∠ACO = θ and ∠ADC = ϕ
∠BCD = ψ and ∠BCO = θ1
In ∆ ACO, tan θ = AO/CO ⇒ tan θ = c/x . . . . . . . . . . . . . . . . . . . . . . . . . (1)
In ∆ ADO, tan ϕ = c/x + d . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Now, θ = α + ϕ (Using ext. ∠ thm.)
⇒ α = θ - ϕ ⇒ tan α = tan (θ – ϕ) ⇒ = tan θ – tan ϕ/1 + tan θ tan ϕ
= c/x – c/x + d / 1 + c/d. c/x + d (using equations (1) and (2)
⇒ tan α = cx + cd – cx/x2 + dx + c2
⇒ x2 + c2 + xd = cd cot α . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
tan ψ = 3c/x +d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (4)
tan θ1 = 3c/x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (5)
But θ1 = ψ + β (by text ∠ thrm)
⇒ β = θ1 – ψ
⇒ tan β = tan (θ1 – ψ) = tan θ1 – tan ψ/ 1+ tan θ1 tan ψ
⇒ tan β = $\frac{\frac{3c}{x}-\frac{3c}{x+d}}{1+\frac{3c}{x}.\frac{3c}{x+d}}$[ Using (4) and (5)]
⇒ tan β = 3cd/x2 + xd + 9c2
⇒ x2 + xd + 9c2 = 3cd cot β . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (6)
From (3) and (6), we get
8c2 = 3cd cot β – cd cot α
⇒ d = 8c/3 cot β – cot α Hence proved.
Thanks
Jitender Pal