\(A=180^0-\left(B+C\right)=60^0\)
Áp dụng định lý hàm sin:
\(\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}=2R\)
\(\Rightarrow\left\{{}\begin{matrix}b=\dfrac{a.sinB}{sinA}=\dfrac{75.sin80^0}{sin60^0}=85,3\left(cm\right)\\c=\dfrac{a.sinC}{sinA}=\dfrac{75.sin40^0}{sin60^0}=55,7\left(cm\right)\\R=\dfrac{a}{2sinA}=\dfrac{75}{2.sin60^0}=25\sqrt{3}\left(cm\right)\end{matrix}\right.\)