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From the Extended Sine Rule , if we denote $R$ to be the radius of circumcircle, we have $R = \dfrac{BC}{2sin(\angle A)} = \dfrac{20}{2sin(50^{\circ})} \approx\boxed{13.054}$ .

Using Sine rule , $R=\dfrac{20}{2\sin 50^\circ} \approx \boxed{13.054}$

You can compute $\sin 50^\circ$ using: $\sin 30^\circ = \sin 150^\circ = 3\sin 50^\circ - 4\sin^3 50^\circ$ . Its tedious and I prefer using calculator :)

Let D be the midpoint on side BC; BD=10 cm; O be the center of the circle.; Let AO= r (radius of circle); Join points A and D.; Now, line AD passes through point O.; AD = AO + OD = r + OD; OD = √[(BO)^2- (BD)^2] = √( r^2 - 10^2); Now in triangle ADB, angle BAD=25°; tan25=BD/AD = 10 /[r + √(r^2 - 10^2)]; Now solving for r, ; r=13.054 cm