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Grade 10Algebra

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.

Hrishant Goswami

12 Years agoGrade 10

1 Answer

Jitender Pal

12 Years ago

Hello Student,

Please find the answer to your question

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)

Again in ∆ ADO

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
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