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If the angle of elevation top of point D of leaning tower at the point A and B is α and β. If slope of tower is θ, then prove that cotθ = (bcotα - αcotβ)/(b - α) Where a and b are the distances of A and B from tower (b > a).

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Let CD is a leaning tower and angles of elevations of top of point D at the place A and B are α and β.

So, ∠DBM = β, ∠DAM = α

Let ∠DCM = θ and DM = h m, CM = x m and AC = a, BC = b.

From right angled ΔDMC,

tan θ = DM/CM

⇒ tan θ = h/x

⇒ x = h/tanθ = h cot θ ….(i)

From right angled ΔDMA,

tan α = DM/AM

tan α = h/(a + x)

⇒ a + x = h/tanα = h cot α

⇒ a = h cot α – x

= h cot α – h cot θ

= h[cot α – cot θ] ……(ii)

For right angled ∆DMB,

tan β = DM/BM

cot β = BM/DM

cot β = (b + x)/h

b + x = h cot β

b = h cot β – x

b = h cot β – h cot θ

= h [cot β – cot θ] ……(iii)

Divide equation (iii) in equation (ii),

a/b = h(cotα - cotθ)/h(cotβ - cotθ)

⇒ a(cot β – cot θ) = b(cot α – cot θ)

⇒ (b – a)cot θ = b cot α – a cot β

⇒ cotθ = (bcotα - αcotβ)/(b - α)

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