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Question Number 88214 by M±th+et£s last updated on 09/Apr/20

Answered by ajfour last updated on 09/Apr/20

$$AB=t,radius=r,AD=DC=b$$$$uppersectionofOB=c$$$$t^2=2b^2;\tan \alpha =\frac{r}{t+r}$$$$\frac{\sqrt{r^2+c^2}}{2b}=\frac{c}{r}=\frac{r}{t+r}$$$$\Rightarrow \frac{\sqrt{r^2+c^2}}{t\sqrt{2}}=\frac{c}{r}=\frac{r}{t+r}$$$$\Rightarrow c=\frac{r^2}{t+r}$$$$\Rightarrow \frac{r^2+{\left(\frac{r^2}{t+r}\right)}^2}{2t^2}=\frac{r^2}{{(t+r)}^2}$$$$lett/r=s$$$$\Rightarrow \frac{1+\frac{1}{{(s+1)}^2}}{2s^2}=\frac{1}{{(s+1)}^2}$$$$\Rightarrow {(s+1)}^2+1=2s^2$$$$s^2-2s-2=0$$$$s=1+\sqrt{3}$$$$\tan \alpha =\frac{1}{s+1}=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}.$$$$\Rightarrow \mathit{\alpha }={\tan }^{-1}(2-\sqrt{3})=15°.$$

Commented by mr W last updated on 09/Apr/20

$$correct!$$

Commented by M±th+et£s last updated on 09/Apr/20

$$nicesolutionsir$$

Answered by mr W last updated on 09/Apr/20

Commented by mr W last updated on 09/Apr/20

$$r=radius$$$$AD=DC$$$$\Rightarrow CF=\frac{OB}{2}=\frac{r}{2}$$$$CD=CE\times \cos \alpha =2r\cos \alpha$$$$CF=CD\times \sin \alpha =2r\cos \alpha \sin \alpha =r\sin 2\alpha$$$$\Rightarrow r\sin 2\alpha =\frac{r}{2}$$$$\Rightarrow \sin 2\alpha =\frac{1}{2}\Rightarrow 2\alpha =30°\Rightarrow \alpha =15°$$

Commented by M±th+et£s last updated on 09/Apr/20

$$thankyousir$$

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