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Question Number 80574 by ajfour last updated on 04/Feb/20

Commented by ajfour last updated on 04/Feb/20

$$Bothcircleshaveequalradius;$$$$Find\alpha .$$

Commented by ajfour last updated on 04/Feb/20

$$a)21.32°b)13.22°c)12.32°d)12.23°$$

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

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

$$OA=\frac{r}{\sin \alpha }$$$$OC=OA+3r=\frac{r}{\sin \alpha }+3r$$$$\frac{CD}{BF}=\frac{AC}{AF}=\frac{3r}{\sqrt{3}r}=\sqrt{3}$$$$\Rightarrow CD=\sqrt{3}r$$$$\frac{CD}{OC}=\tan \alpha$$$$\frac{\sqrt{3}r}{\frac{r}{\sin \alpha }+3r}=\tan \alpha$$$$\frac{\sqrt{3}}{\frac{1}{\sin \alpha }+3}=\frac{\sin \alpha }{\cos \alpha }$$$$\frac{\sqrt{3}}{1+3\sin \alpha }=\frac{1}{\cos \alpha }$$$$\sqrt{3}\cos \alpha -3\sin \alpha =1$$$$\frac{1}{2}\cos \alpha -\frac{\sqrt{3}}{2}\sin \alpha =\frac{\sqrt{3}}{6}$$$$\sin (\frac{\pi }{6}-\alpha )=\frac{\sqrt{3}}{6}$$$$\Rightarrow \alpha =\frac{\pi }{6}-{\sin }^{-1}\frac{\sqrt{3}}{6}\approx 13.22°$$

Commented by ajfour last updated on 04/Feb/20

$$ThankyouSir.$$

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