Question Number 57328 by ANTARES VY last updated on 02/Apr/19
Answered by mr W last updated on 02/Apr/19
$${let}\:\alpha=\frac{\angle{A}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\alpha=\frac{{r}}{{AO}}=\frac{\mathrm{2}×\mathrm{3}}{\mathrm{10}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\frac{\mathrm{3}}{\sqrt{\mathrm{5}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${BC}=\mathrm{2}×\left({AO}+{r}\right)\:\mathrm{tan}\:\alpha \\ $$$$=\mathrm{2}×\left(\frac{\mathrm{10}}{\mathrm{3}}+\mathrm{2}\right)×\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$=\mathrm{8} \\ $$