$tan\widehat{B}=\frac{AC}{AB}<=>tan40^o=\frac{AC}{12}=>AC=tan40^o.12\approx10,07(cm)$ Áp dụng định lí Pytago,ta có: $AB^2+AC^2=BC^2<=>12^2+10,07^2=BC^2$ $<=>BC^2=245,4049<=>BC\approx15,67(cm)$
b.
Ta có: Áp dụng định lí Pytago,ta có: $AB^2+AC^2=BC^2<=>5^2+6^2=BC^2$ $<=>BC^2=61<=>BC\approx7,81(cm)$