đặt : \(AB=c\) ; \(AC=b\) ; \(BC=a\) và \(AH=h_a\)
ta có : \(\widehat{A}=180-\left(\widehat{B}+\widehat{C}\right)=180-\left(30+40\right)=110\)
ta lại có : \(\dfrac{b}{sinB}=\dfrac{a}{sinA}\Leftrightarrow\dfrac{12}{sin30}=\dfrac{a}{sin110}\Leftrightarrow a=\dfrac{12}{sin30}.sin110\)
\(\Leftrightarrow a\simeq22,55\)
ta có : \(S_{\Delta ABC}=\dfrac{1}{2}a.b.sinC=\dfrac{1}{2}.\left(22,55\right).12.sin40\simeq86,97\)
ta có : \(S_{\Delta ABC}=\dfrac{1}{2}a.h_a\Leftrightarrow86,97=\dfrac{1}{2}.\left(22,55\right).h_a\Leftrightarrow h_a=86,97.\dfrac{2}{22,55}\simeq7,71\)
vậy \(AH=7,71\) và \(BC=22,55\)