Vẽ đường cao CH. Ta có:
\(\hept{\begin{cases}BH+AH=14\\BH^2+CH^2=225\\AH^2+CH^2=169\end{cases}\Rightarrow\hept{\begin{cases}BH+AH=14\\BH^2-AH^2=56\end{cases}\Leftrightarrow}\hept{\begin{cases}BH+AH=14\\\left(BH+AH\right)\left(BH-AH\right)=56\end{cases}\Leftrightarrow}\hept{\begin{cases}BH+AH=14\\BH-AH=4\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}BH=9\\AH=5\end{cases}\Rightarrow\hept{\begin{cases}\widehat{A}=cos^{-1}\frac{5}{13}\approx67^023'\\\widehat{B}=cos^{-1}\frac{9}{15}\approx53^08'\\\widehat{C}\approx180^0-\left(67^023'+53^08'\right)=59^029'\end{cases}}}\)
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