In a ∆ABC, ∠B = 60°; ∠C = 40°. If I is the incentre, then find ∠BIC.

ΔABC with ∠B = $$60^\circ$$​ and $$\angle C$$​ = $$40^\circ.$$​​

I is the incentre of the triangle.

The angle ∠BIC is calculated using the formula for the angle formed by the incentre with two vertices of a triangle:

​$$\angle BIC = 90^\circ + \frac{A}{2}$$​​

where A is the third angle of the triangle.

The sum of the interior angles of a triangle is 180°:

Solution: 

​$$\angle A = 180^\circ - (60^\circ + 40^\circ) = 180^\circ - 100^\circ = 80^\circ$$​​

Now, using the formula for ∠BIC:

​$$\angle BIC = 90^\circ + \frac{80^\circ}{2} = 90^\circ + 40^\circ = 130^\circ.$$​​

Thus, $$\bf \angle BIC = 130^\circ.$$​​