Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos B = \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\\
\Leftrightarrow \cos 45^\circ = \frac{{{{15}^2} + {c^2} - {{\left( {15\sqrt 2 } \right)}^2}}}{{2.15.c}}\\
\Leftrightarrow 15\sqrt 2 c = {c^2} - 225\\
\Leftrightarrow {c^2} - 15\sqrt 2 c - 225 = 0\\
\Rightarrow c = \frac{{15.\left( {\sqrt 6 + \sqrt 2 } \right)}}{2}\\
\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R\\
\Leftrightarrow \frac{{15}}{{\sin A}} = \frac{{15\sqrt 2 }}{{\sin 45^\circ }} = \frac{c}{{\sin C}}\\
\Rightarrow \sin A = \frac{{15.\sin 45^\circ }}{{15\sqrt 2 }} = \frac{1}{2} \Rightarrow \widehat A = 30^\circ \\
\widehat C = 180^\circ - \widehat A - \widehat C = 105^\circ
\end{array}\)
