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Question Number 74509 by ajfour last updated on 25/Nov/19

Commented by ajfour last updated on 25/Nov/19

$$Findtheradiusofcircle.△ABC$$$$isequilateralwithsideunity.$$

Commented by ajfour last updated on 25/Nov/19

Commented by ajfour last updated on 25/Nov/19

$$AT=r\sqrt{3},\sin \alpha =\frac{r}{1-r\sqrt{3}}$$$$\cos (60°-\alpha )=BF$$$$\frac{\cos \alpha +\sqrt{3}\sin \alpha }{2}=(1-r\sqrt{3})\cos \alpha$$$$\Rightarrow 1+\sqrt{3}\tan \alpha =2-2r\sqrt{3}$$$$\Rightarrow \frac{r\sqrt{3}}{\sqrt{{(1-r\sqrt{3})}^2-r^2}}=1-2r\sqrt{3}$$$$\Rightarrow 3r^2={(1-2r\sqrt{3})}^2(1-2r\sqrt{3}+2r^2)$$$$\Rightarrow -3r^2+1-2r\sqrt{3}+2r^2+12r^2$$$$-24\sqrt{3}{r}^3+24r^4-4\sqrt{3}r+24r^2$$$$-8\sqrt{3}{r}^3=0\Rightarrow$$$$24r^4-32\sqrt{3}{r}^3+35r^2-6\sqrt{3}r+1=0$$$$possiblyr\approx 0.16739(\mathrm{\&}1.5293)$$

Commented by ajfour last updated on 25/Nov/19

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