\(cos\widehat{ACB}=\frac{AC^2+CB^2-AB^2}{2.AC.CB}=\frac{13^2+14^2-15^2}{2.13.14}=\frac{5}{13}\)
Xét ΔAHC vuông tại H
\(CH=cos\widehat{ACB}.AC=\frac{5}{13}.13=5cm\)
Cách khác:
Ta có:\(AB^2=AH^2+HB^2\)
Hay:\(AB^2=AC^2-HC^2+\left(BC-HC\right)^2\)
\(\Leftrightarrow225=169-HC^2+196-28HC+HC^2\)
\(\Leftrightarrow365-28HC=225\)
\(\Leftrightarrow28HC=140\)
\(\Leftrightarrow HC=5\)(cm)