Trong △ABH vuông tại H có:
\(sin\widehat{B}=sin60^0=\dfrac{AH}{AB}=\dfrac{\sqrt{3}}{2}\)
\(\Rightarrow AB=\dfrac{2\sqrt{3}}{3}.AH=\dfrac{2\sqrt{3}}{3}.12=8\sqrt{3}\).
Mặt khác: \(tn\widehat{B}=tn60^0=\dfrac{AH}{BH}=\sqrt{3}\Rightarrow BH=\dfrac{AH\sqrt{3}}{3}=\dfrac{12\sqrt{3}}{3}=4\sqrt{3}\).
\(S_{AHC}=3S_{AHB};S_{AHC}+S_{AHB}=S_{ABC}\Rightarrow S_{AHB}=\dfrac{S_{ABC}}{4}\).
\(\dfrac{S_{AHB}}{S_{ABC}}=\dfrac{HB}{BC}=\dfrac{1}{4}\Rightarrow BC=4HB=4.4\sqrt{3}=16\sqrt{3}\).
\(\Rightarrow HC=BC-HB=16\sqrt{3}-4\sqrt{3}=12\sqrt{3}\).
△AHC vuông tại H có:
\(tn\widehat{ACH}=\dfrac{AH}{CH}=\dfrac{12}{12\sqrt{3}}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}\).
\(\Rightarrow\widehat{ACB}=30^0\).
\(\Rightarrow\widehat{BAC}=180^0-\widehat{ABC}-\widehat{ACB}=180^0-60^0=30^0=90^0\).
\(\Rightarrow\)△ABC vuông tại A.
\(\Rightarrow AB^2+AC^2=BC^2\Rightarrow AC=\sqrt{BC^2-AB^2}=\sqrt{16^2.3-8^2.3}=24\)
