I is the incentre of ∆ABC, ∠ABC = 60° and ∠ACB

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On the basis of given in question , we draw a figure triangle ABC ,

∠ABC = 60°, ACB = 50°

∠IBC =	1	∠ABC = 30°
	2

∠ICB =	1	∠ACB = 25°
	2

We know that , ∠BIC + ∠ICB + ∠IBC = 180°
∴ ∠BIC = 180° – 30° – 25° = 125°