Giải thích các bước giải:
a,
Ta có:
\(\begin{array}{l}
\cos C = \frac{{{a^2} + {b^2} - {c^2}}}{{2.a.b}} = \frac{{{8^2} + {{10}^2} - {{13}^2}}}{{2.8.10}} = - \frac{1}{{32}}\\
\cos C < 0 \Rightarrow 90^\circ < \widehat C < 180^\circ
\end{array}\)
Do đó, tam giác ABC có góc C là góc tù.
b,
\(\begin{array}{l}
\cos \widehat C = - \frac{1}{{32}} \Rightarrow \sin \widehat C = \sqrt {1 - {{\cos }^2}C} = \frac{{\sqrt {1023} }}{{32}}\\
\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R\\
\Rightarrow R = \frac{c}{{2\sin C}} = \frac{{13}}{{\frac{{2.\sqrt {1023} }}{{32}}}} = \frac{{208}}{{\sqrt {1023} }}\\
c,\\
{S_{ABC}} = \frac{{abc}}{{4R}} = \frac{{8.10.13}}{{4.\frac{{208}}{{\sqrt {1023} }}}} = \frac{{5\sqrt {1023} }}{4}
\end{array}\)
