giải giúp mik

a) Áp dụng định lí Cosin trong $\displaystyle \vartriangle ABC$ có:
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
BC^{2} =AB^{2} +AC^{2} -2AB.AC.\cos A\\
=1^{2} +2^{2} -2.2.1.\cos 120^{0} =7\\
\Rightarrow BC=\sqrt{7}\\
\cos C=\frac{AC^{2} +BC^{2} -AB^{2}}{2.CA.CB} =\frac{2^{2} +\left(\sqrt{7}\right)^{2} -1^{2}}{2.2.\sqrt{7}} =\frac{5\sqrt{7}}{14}
\end{array}$
b) Đặt $\displaystyle AD=x$
$\displaystyle \widehat{DAB} +\widehat{CAB} =180^{0} \Rightarrow \widehat{DAB} =180^{0} -120^{0} =60^{0}$
Áp dụng định lí Cosin trong $\displaystyle \vartriangle ABD$ có:
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
BD^{2} =AB^{2} +AD^{2} -2AB.AD.\cos A\\
\Rightarrow 2^{2} =1^{2} +x^{2} -2.1.x.\cos 60^{0}\\
\Rightarrow x^{2} -x-3=0\\
\Rightarrow x^{2} -x+\frac{1}{4} =\frac{13}{4}\\
\Rightarrow \left( x-\frac{1}{2}\right)^{2} =\frac{13}{4}\\
\Rightarrow \left[ \begin{array}{l l}
x & =\frac{1+\sqrt{13}}{2}\\
x & =\frac{1-\sqrt{13}}{2} \ ( vl)
\end{array} \right. \Rightarrow AD=\frac{1+\sqrt{13}}{2}
\end{array}$