Question Number 96785 by M±th+et+s last updated on 04/Jun/20
Answered by mr W last updated on 04/Jun/20
Commented by mr W last updated on 04/Jun/20
$${r}+{r}\:\mathrm{cos}\:\alpha={a} \\ $$$$\Rightarrow\mathrm{cos}\:\alpha=\frac{{a}}{{r}}−\mathrm{1}=\frac{\mathrm{144}}{\mathrm{80}}−\mathrm{1}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{tan}\:\beta=\frac{{r}}{{a}−{r}\:\mathrm{sin}\:\alpha}=\frac{\mathrm{1}}{\frac{{a}}{{r}}−\mathrm{sin}\:\alpha}=\frac{\mathrm{1}}{\frac{\mathrm{144}}{\mathrm{80}}−\frac{\mathrm{3}}{\mathrm{5}}}=\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$\gamma=\frac{\pi}{\mathrm{2}}−\mathrm{2}\beta \\ $$$${m}=\frac{{a}}{\mathrm{cos}\:\gamma}=\frac{{a}}{\mathrm{sin}\:\mathrm{2}\beta}=\frac{{a}}{\mathrm{2}\:\mathrm{sin}\:\beta\:\mathrm{cos}\:\beta} \\ $$$$=\frac{\mathrm{144}}{\mathrm{2}×\frac{\mathrm{5}×\mathrm{6}}{\mathrm{5}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} }}=\frac{\mathrm{732}}{\mathrm{5}}=\mathrm{146}.\mathrm{4} \\ $$
Commented by M±th+et+s last updated on 04/Jun/20
$${great}\:{work}\:{mr}\:{W}.\:{thanx} \\ $$