Q.39. Trigonometry, Properties of Triangles

Question 39.
Let ABC a traingle with AB = AC and I be a incentre with BC = AB + AI.  
Find angle BAC.

Answer 39.
Let AB = AC = x
and ∠ BAC = 2θ
=> BC = 2xsinθ

Using the property of triangle,
rs = Δ, where
r = in-radius,
s = semiperimeter = (1/2)(AB + AC + BC) = (1/2)(x + x + 2xsinθ) = x + xsinθ
Δ = (1/2)x^2 * sin2θ = x^2 sinθcosθ
=> r * (x + xsinθ) = x^2 sinθcosθ
=> r/x = sinθcosθ/(1 + sinθ)

Using the given condition,
BC = AB + AI
=> BC = AB + r/sinθ
=> 2xsinθ = x + xcosθ/(1 + sinθ)
=> 2sinθ(1 + sinθ) = 1 + sinθ + cosθ
=> 2sinθ + 2sin^2 θ = 1 + sinθ + cosθ
=> sinθ - cosθ + sin^2 θ + 1 - cos^2 θ = 1
=> sinθ - cosθ + sin^2 θ - cos^2 θ = 0
=> (sinθ - cosθ) (1 + sinθ + cosθ) = 0
=> sinθ - cosθ = 0 or 1 + sinθ + cosθ = 0
sinθ - cosθ = 0 => tanθ = 1 => θ = 45°
[second equation does not give θ < 90°]
=> ∠ BAC = 2θ = 2*45 = 90°.

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