Đáp án:
$\begin{array}{l}
Theo\,Pytago:\\
B{C^2} = A{B^2} + A{C^2}\\
\Rightarrow A{C^2} = {13^2} - {10^2} = 69\\
\Rightarrow AC = \sqrt {69} \left( {cm} \right)\\
+ \sin \widehat B = \dfrac{{AC}}{{BC}} = \dfrac{{\sqrt {69} }}{{13}} = {40^0}\\
\Rightarrow \widehat C = {50^0}\\
Vay\,AC = \sqrt {69} cm;\widehat B = {40^0};\widehat C = {50^0}
\end{array}$