Menu Close

Question-80574




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

Leave a Reply

Your email address will not be published. Required fields are marked *