In a triangle $$ABC$$, $$\angle A = 40^\circ$$, $$\angle B = 60^\circ$$. If the bisector of angle $$A$$ meets the circum circle of triangle $$ABC$$ in $$X$$ then the angle subtended by the segment $$BX$$ at the circum-centre is

Solution

Since $$AX$$ bisects $$\angle A$$, $$\angle BAX = 20^\circ$$. This is an inscribed angle standing on the chord $$BX$$, and the angle at the centre is twice the inscribed angle on the same arc. So the required angle is $$2 \times 20^\circ = 40^\circ$$.

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