I is the incentre of a triangle ABC. If ∠ACB = 55°, ∠ABC = 65° then the value of ∠BIC is

According to question,
Given :
∠ACB = 55°
∠ABC = 65°
∠BIC = ?

∴ ∠ACB + ∠ABC + ∠BAC = 180°
∠BAC = 180° - 55° - 65°
∠BAC = 60°
We know that
∠BIC = 90 + $$\frac{1}{2}$$ ∠A
∠BIC = 90 + $$\frac{1}{2}$$ × 60
∠BIC = 90 + 30
∠BIC = 120°

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Alternate:
In ΔBIC,
$$\frac{1}{2}$$ ∠B + $$\frac{1}{2}$$ ∠C + ∠BIC = 180°
$$\frac{1}{2}$$ (65° + 55°) + ∠BIC = 180°
∠BIC = 180° - 60°
∠BIC = 120°