Question Number 80574 by ajfour last updated on 04/Feb/20
Commented by ajfour last updated on 04/Feb/20
$${Both}\:{circles}\:{have}\:{equal}\:{radius}; \\ $$$${Find}\:\alpha. \\ $$
Commented by ajfour last updated on 04/Feb/20
$$\left.{a}\left.\right)\left.\:\left.\mathrm{21}.\mathrm{32}°\:\:\:{b}\right)\mathrm{13}.\mathrm{22}°\:\:{c}\right)\:\mathrm{12}.\mathrm{32}°\:\:{d}\right)\:\:\mathrm{12}.\mathrm{23}° \\ $$
Answered by mr W last updated on 04/Feb/20
Commented by mr W last updated on 04/Feb/20
$${OA}=\frac{{r}}{\mathrm{sin}\:\alpha} \\ $$$${OC}={OA}+\mathrm{3}{r}=\frac{{r}}{\mathrm{sin}\:\alpha}+\mathrm{3}{r} \\ $$$$\frac{{CD}}{{BF}}=\frac{{AC}}{{AF}}=\frac{\mathrm{3}{r}}{\:\sqrt{\mathrm{3}}{r}}=\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{CD}=\sqrt{\mathrm{3}}{r} \\ $$$$\frac{{CD}}{{OC}}=\mathrm{tan}\:\alpha \\ $$$$\frac{\sqrt{\mathrm{3}}{r}}{\frac{{r}}{\mathrm{sin}\:\alpha}+\mathrm{3}{r}}=\mathrm{tan}\:\alpha \\ $$$$\frac{\sqrt{\mathrm{3}}}{\frac{\mathrm{1}}{\mathrm{sin}\:\alpha}+\mathrm{3}}=\frac{\mathrm{sin}\:\alpha}{\mathrm{cos}\:\alpha} \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{1}+\mathrm{3}\:\mathrm{sin}\:\alpha}=\frac{\mathrm{1}}{\mathrm{cos}\:\alpha} \\ $$$$\sqrt{\mathrm{3}}\:\mathrm{cos}\:\alpha−\mathrm{3}\:\mathrm{sin}\:\alpha=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{cos}\:\alpha−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{sin}\:\alpha=\frac{\sqrt{\mathrm{3}}}{\mathrm{6}} \\ $$$$\mathrm{sin}\:\left(\frac{\pi}{\mathrm{6}}−\alpha\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{6}} \\ $$$$\Rightarrow\alpha=\frac{\pi}{\mathrm{6}}−\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}}{\mathrm{6}}\approx\mathrm{13}.\mathrm{22}° \\ $$
Commented by ajfour last updated on 04/Feb/20
$${Thank}\:{you}\:{Sir}. \\ $$