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
\(1\)
\(\sqrt {11} \)
\(3\sqrt 3 \)
\(10\)