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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 α.

$${Both}\:{circles}\:{have}\:{equal}\:{radius}; \\ $$$${Find}\:\alpha. \\ $$

Commented by ajfour last updated on 04/Feb/20

a) 21.32° b)13.22° c) 12.32° d) 12.23°

$$\left.{a}\left.\right)\left.\:\left.\mathrm{21}.\mathrm{32}°\:\:\:{b}\right)\mathrm{13}.\mathrm{22}°\:\:{c}\right)\:\mathrm{12}.\mathrm{32}°\:\:{d}\right)\:\:\mathrm{12}.\mathrm{23}° \\ $$

Answered by mr W last updated on 04/Feb/20

Commented by mr W last updated on 04/Feb/20

OA=(r/(sin α)) OC=OA+3r=(r/(sin α))+3r ((CD)/(BF))=((AC)/(AF))=((3r)/((√3)r))=(√3) ⇒CD=(√3)r ((CD)/(OC))=tan α (((√3)r)/((r/(sin α))+3r))=tan α ((√3)/((1/(sin α))+3))=((sin α)/(cos α)) ((√3)/(1+3 sin α))=(1/(cos α)) (√3) cos α−3 sin α=1 (1/2) cos α−((√3)/2) sin α=((√3)/6) sin ((π/6)−α)=((√3)/6) ⇒α=(π/6)−sin^(−1) ((√3)/6)≈13.22°

$${OA}=\frac{{r}}{\mathrm{sin}\:\alpha} \\ $$$${OC}={OA}+\mathrm{3}{r}=\frac{{r}}{\mathrm{sin}\:\alpha}+\mathrm{3}{r} \\ $$$$\frac{{CD}}{{BF}}=\frac{{AC}}{{AF}}=\frac{\mathrm{3}{r}}{\sqrt{\mathrm{3}}{r}}=\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{CD}=\sqrt{\mathrm{3}}{r} \\ $$$$\frac{{CD}}{{OC}}=\mathrm{tan}\:\alpha \\ $$$$\frac{\sqrt{\mathrm{3}}{r}}{\frac{{r}}{\mathrm{sin}\:\alpha}+\mathrm{3}{r}}=\mathrm{tan}\:\alpha \\ $$$$\frac{\sqrt{\mathrm{3}}}{\frac{\mathrm{1}}{\mathrm{sin}\:\alpha}+\mathrm{3}}=\frac{\mathrm{sin}\:\alpha}{\mathrm{cos}\:\alpha} \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{1}+\mathrm{3}\:\mathrm{sin}\:\alpha}=\frac{\mathrm{1}}{\mathrm{cos}\:\alpha} \\ $$$$\sqrt{\mathrm{3}}\:\mathrm{cos}\:\alpha−\mathrm{3}\:\mathrm{sin}\:\alpha=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{cos}\:\alpha−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{sin}\:\alpha=\frac{\sqrt{\mathrm{3}}}{\mathrm{6}} \\ $$$$\mathrm{sin}\:\left(\frac{\pi}{\mathrm{6}}−\alpha\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{6}} \\ $$$$\Rightarrow\alpha=\frac{\pi}{\mathrm{6}}−\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}}{\mathrm{6}}\approx\mathrm{13}.\mathrm{22}° \\ $$

Commented by ajfour last updated on 04/Feb/20

Thank you Sir.

$${Thank}\:{you}\:{Sir}. \\ $$

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