In the adjoining figure, $$\angle ACD=38^\circ$$. Then the measure of angle $$x$$ in degrees is

The equal side marks give $$AD=CD$$, so $$\angle CAD=\angle ACD=38^\circ$$ and $$\angle ADC=104^\circ$$. Since $$ABCD$$ is cyclic, $$\angle ABC=180^\circ-104^\circ=76^\circ$$. The other equal side marks give $$AB=AC$$, so $$\angle BCA=76^\circ$$. The exterior angle at $$A$$ equals the opposite interior angle $$\angle BCD=76^\circ+38^\circ=114^\circ$$.