Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
B{C^2} = A{B^2} + A{C^2} - 2AB.AC.\cos A\\
\Leftrightarrow B{C^2} = {10^2} + {4^2} - 2.4.10.\cos 60^\circ \\
\Leftrightarrow B{C^2} = 76\\
\Rightarrow BC = 2\sqrt {19} \\
p = \frac{{AB + AC + BC}}{2} = \frac{{4 + 10 + 2\sqrt {19} }}{2} = 7 + \sqrt {19} \\
S = \sqrt {p\left( {p - AB} \right)\left( {p - AC} \right)\left( {p - BC} \right)} = 10\sqrt 3 \\
S = \frac{1}{2}{h_b}.AC \Leftrightarrow 10\sqrt 3 = \frac{1}{2}.{h_b}.4 \Leftrightarrow {h_b} = 5\sqrt 3 \\
\sin C = \frac{{{h_b}}}{{BC}} = \frac{{5\sqrt 3 }}{{2\sqrt {19} }} \Rightarrow \tan C = .....
\end{array}\)
