Question Number 150609 by ajfour last updated on 13/Aug/21
$${I}\:{thought}\:{this}\:{as}\:{more}\:{basic}: \\ $$$$\frac{{sinA}}{{a}}=\frac{\mathrm{1}}{\mathrm{2}{R}} \\ $$$$\frac{{cosA}}{{a}}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{abc}} \\ $$$$\Rightarrow\:\:\boldsymbol{{tanA}}=\frac{\boldsymbol{{abc}}}{\boldsymbol{{R}}\left(\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} \right)} \\ $$