Ambiguous case and circumradius

Problem

In a triangle $ABC$ it is known that $\overline{BC}=14$, $\beta=A\hat{B}C=12{,}37^\circ$, $\overline{AC}=7$. a) Find the missing elements with the sine and cosine rules, showing that the angle $\gamma=A\hat{C}B$ (and hence side $\overline{AB}$) has more than one solution. b) Choosing the solution with $\gamma=142{,}27^\circ$, find the circumradius with the chord theorem.

Solution

a) By the sine rule ($a=\overline{BC}=14$ opposite $\alpha$, $b=\overline{AC}=7$ opposite $\beta$): $\sin\alpha=\frac{a\sin\beta}{b}=\frac{14\sin 12{,}37^\circ}{7}=2\sin 12{,}37^\circ\approx0{,}4283\ \Rightarrow\ \alpha\approx25{,}36^\circ\ \lor\ \alpha\approx154{,}64^\circ.$ First solution: $\alpha=25{,}36^\circ\Rightarrow\gamma=180^\circ-\beta-\alpha=142{,}27^\circ$, and $c=\overline{AB}=\dfrac{b\sin\gamma}{\sin\beta}\approx20$. Second solution: $\alpha=154{,}64^\circ\Rightarrow\gamma\approx12{,}99^\circ$, $\overline{AB}\approx7{,}35$. So the problem admits two triangles. b) Chord/sine rule: $2R=\dfrac{c}{\sin\gamma}=\dfrac{20}{\sin 142{,}27^\circ}\approx32{,}69$, hence $R\approx16{,}34$. $\boxed{\,\gamma\approx142{,}27^\circ,\quad \overline{AB}\approx20,\quad R\approx16{,}34\,}$