Quiz 5:正弦定律與外接圓半徑

於 \(\vartriangle ABC\) ,若 \(\overline {BC}  = 10\) , \(\angle BAC = \frac{\pi }{6}\) , \(\angle BCA = \frac{\pi }{4}\) ,則 \(\vartriangle ABC\) 之外接圓半徑為多少

\(1\)

\(\sqrt {11} \)

\(3\sqrt 3 \)

\(10\)