\(A=180^0-\left(B+C\right)=63^0\)
Áp dụng định lý hàm sin:
\(\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}\)
\(\Rightarrow\left\{{}\begin{matrix}b=\dfrac{a.sinB}{sinA}=\dfrac{8.sin47^0}{sin63^0}\approx6,57\left(cm\right)\\c=\dfrac{a.sinC}{sinA}\approx8,44\left(cm\right)\end{matrix}\right.\)