\(A+B+C=180^o\Rightarrow A=180^o-\left(B+C\right)=180^o-\left(60^o+40^o\right)=80^o\)
Gọi \(BC=a;AC=b;AB=c\)
Theo định lý sin cho tam giác :
\(\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}\)
\(\Rightarrow\left\{{}\begin{matrix}b=AC=\dfrac{a.sinB}{sinA}\approx\dfrac{6.0,9}{1}=5,4\left(cm\right)\\c=AB=\dfrac{a.sinC}{sinA}\approx\dfrac{6.0,6}{1}=3,6\left(cm\right)\end{matrix}\right.\)