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Question Number 159911 by amin96 last updated on 22/Nov/21

Answered by mr W last updated on 22/Nov/21

Commented by mr W last updated on 22/Nov/21

$$assumeABCDissquarewithside$$$$lengtha.$$$$ED=a\tan \alpha$$$$\alpha =\frac{\pi }{6}-\theta$$$$AE=a(1-\tan \alpha )$$$$\sin \frac{\theta }{2}=\frac{r_1-r_2}{r_1+r_2}=\frac{2r_1}{r_1+r_2}-1$$$$\Rightarrow r_1+r_2=\frac{2r_1}{1+\sin \frac{\theta }{2}}$$$$r_1+r_2+(r_1+r_2)\cos \frac{\theta }{2}=a$$$$(r_1+r_2)(1+\cos \frac{\theta }{2})=a$$$$\frac{2r_1}{1+\sin \frac{\theta }{2}}(1+\cos \frac{\theta }{2})=a$$$$\Rightarrow \frac{r_1}{a}=\frac{1+\sin \frac{\theta }{2}}{2(1+\cos \frac{\theta }{2})}$$$$AE=r_1+\frac{r_1}{\tan (\frac{\pi }{4}-\frac{\theta }{2})}=a(1-\tan \alpha )$$$$\frac{r_1}{a}(1+\frac{1}{\tan (\frac{\pi }{4}-\frac{\theta }{2})})=1-\tan (\frac{\pi }{6}-\theta )$$$$\frac{1+\sin \frac{\theta }{2}}{2(1+\cos \frac{\theta }{2})}(1+\frac{1}{\tan (\frac{\pi }{4}-\frac{\theta }{2})})=1-\tan (\frac{\pi }{6}-\theta )$$$$$$$$\frac{(1+\sin \frac{\theta }{2})\cos \frac{\theta }{2}}{(1+\cos \frac{\theta }{2})(\cos \frac{\theta }{2}-\sin \frac{\theta }{2})}=1-\tan (\frac{\pi }{6}-\theta )$$$$\Rightarrow \theta \approx 6.0796°$$$$\sin \frac{\theta }{2}=\frac{\frac{r_1}{r_2}-1}{\frac{r_1}{r_2}+1}$$$$\Rightarrow \frac{r_1}{r_2}=\frac{1+\sin \frac{\theta }{2}}{1-\sin \frac{\theta }{2}}$$$$\Rightarrow \frac{S_1}{S_2}={\left(\frac{r_1}{r_2}\right)}^2={\left(\frac{1+\sin \frac{\theta }{2}}{1-\sin \frac{\theta }{2}}\right)}^2\approx 1.53$$

Commented by Tawa11 last updated on 22/Nov/21

$$\mathrm{Weldone}\mathrm{sir}$$

Commented by mr W last updated on 22/Nov/21

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