Giải thích các bước giải:
\(\begin{array}{l}
12,\\
\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2.b.c}} = \frac{{{{13}^2} + {{24}^2} - {{15}^2}}}{{2.13.24}} = \frac{5}{6}\\
\Rightarrow \sin A = \sqrt {1 - {{\cos }^2}A} = \frac{{\sqrt {11} }}{6}\\
{S_{ABC}} = \frac{1}{2}.bc.\sin A = \frac{1}{2}.13.24.\frac{{\sqrt {11} }}{6} = 26\sqrt {11} \,\left( {c{m^2}} \right)\\
13,\\
\overrightarrow {AB} = \left( {1;7} \right)\,\,\,\,\, \Rightarrow AB = \sqrt {{1^2} + {7^2}} = 5\sqrt 2 \\
\overrightarrow {AC} = \left( { - 6;8} \right)\,\,\,\, \Rightarrow AC = \sqrt {{{\left( { - 6} \right)}^2} + {8^2}} = 10\\
\overrightarrow {BC} = \left( { - 7;1} \right) \Rightarrow BC = \sqrt {{{\left( { - 7} \right)}^2} + {1^2}} = 5\sqrt 2 \\
A{B^2} + B{C^2} = 100 = A{C^2};\,\,\,\,\,AB = BC
\end{array}\)
Suy ra tam giác ABC vuông cân tại B.
\(\begin{array}{l}
\overrightarrow {AC} = \left( { - 6;8} \right) \Rightarrow \overrightarrow {CA} = \left( {6; - 8} \right)\\
\overrightarrow {AB} = \left( {1;7} \right)\\
\Rightarrow \overrightarrow {AB} .\overrightarrow {CA} = 6.1 + \left( { - 8} \right).7 = - 50
\end{array}\)
