In the adjoining figure, $$\angle ABC=60^\circ$$ and $$\angle ACB=80^\circ$$. $$AD$$ bisects $$\angle A$$. Through $$C$$, a line making an angle of $$\frac{\angle A}{2}$$ with $$BC$$ is drawn. It intersects the bisector and the perpendicular from $$B$$ to $$AD$$ as shown. The value of $$x$$ in degrees is
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Since the angles at $$B$$ and $$C$$ are $$60^\circ$$ and $$80^\circ$$, we have $$\angle A=40^\circ$$, so each half is $$20^\circ$$. The bisector line makes $$80^\circ$$ with $$BC$$, and its perpendicular through $$B$$ makes $$10^\circ$$ with $$BC$$ on the other side. The required exterior angle is therefore $$10^\circ+20^\circ=30^\circ$$.