Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} \Rightarrow {a^2} = {b^2} + {c^2} - 2bc.\cos A\\
\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R\\
2,\\
{S_{ABC}} = \frac{1}{2}.bc.\sin A = \frac{1}{2}ab.\sin C = \frac{1}{2}ac.\sin B\\
3,\\
\overrightarrow a .\overrightarrow b = 2.4 + 3.\left( { - 1} \right) = 5\\
4,\\
{S_{ABC}} = \frac{1}{2}bc.\sin A = \frac{1}{2}.10.20.sin60^\circ = 50\sqrt 3 \\
5,\\
p = \frac{{a + b + c}}{2} = 10\\
\Rightarrow {S_{ABC}} = \sqrt {p\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right)} = 10\sqrt 3 \\
6,\\
p = \frac{{a + b + c}}{2} = \frac{{5 + 3 + 4}}{2} = 6
\end{array}\)
