góc A=180-50-60=70 độ
Xét ΔABC có BC/sinA=AB/sinC=AC/sin B
=>BC/sin70=12/sin60=AC/sin50
=>\(BC\simeq13,02;AC\simeq10,61\)
\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinBAC=\dfrac{1}{2}\cdot12\cdot10.61\cdot sin70\simeq59,82\)
\(AH=2\cdot\dfrac{59.82}{10.61}\simeq11,28\)
\(HB=\sqrt{AB^2-AH^2}=\sqrt{12^2-11.28^2}\simeq4,09\)
HC=10,61-4,09=6,52