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Question Number 213173 by ajfour last updated on 31/Oct/24

Commented by ajfour last updated on 31/Oct/24

$$Findr$$

Answered by ajfour last updated on 31/Oct/24

Commented by ajfour last updated on 01/Nov/24

$$\varphi =\frac{90°-2\theta }{2}=45°-\theta$$$$r\tan \varphi =r\left(\frac{1-\tan \theta }{1+\tan \theta }\right)=1-2r$$$$\Rightarrow \tan \theta =\frac{r-(1-2r)}{r+(1-2r)}=\frac{3r-1}{1-r}$$$${(x+r)}^2+1={(x+1-r)}^2$$$$\Rightarrow 2rx=2x-2r-2rx$$$$x=\frac{r}{1-2r}$$$$Further$$$$\frac{r}{x}=\tan \theta =\frac{3r-1}{1-r}$$$$\Rightarrow x=\frac{r(1-r)}{3r-1}=\frac{r}{1-2r}$$$$\Rightarrow (1-r)(1-2r)=3r-1$$$$\Rightarrow 2r^2-6r+2=0$$$$\Rightarrow r^2-3r+1=0$$$$r=\frac{3}{2}-\sqrt{\frac{9}{4}-1}$$$$r=\frac{3-\sqrt{5}}{2}$$

Commented by Frix last updated on 31/Oct/24

$$\mathrm{Your}\mathrm{answer}\mathrm{is}\mathrm{right}.$$

Commented by ajfour last updated on 01/Nov/24

Commented by ajfour last updated on 01/Nov/24

Thank you (all concerned).

Commented by ajfour last updated on 01/Nov/24

$$\tan 2\theta comesoutequalto2.$$

Answered by a.lgnaoui last updated on 31/Oct/24

$$\tan \mathrm{x}=\frac{\mathrm{r}}{2\mathrm{r}}=\frac{1}{2}=\frac{2\mathrm{t}}{1-{\mathrm{t}}^2}\mathrm{t}=\tan \frac{\mathrm{x}}{2}$$$${\mathrm{t}}^2+4\mathrm{t}-1=0\mathrm{t}=\frac{-4\pm 2\sqrt{5}}{2}=\sqrt{5}-2$$$$$$$$\tan \left(\frac{90-\mathrm{x}}{2}\right)=\frac{\mathrm{r}}{1-\mathrm{r}}$$$$\frac{\tan 45-\tan \frac{\mathrm{x}}{2}}{1+\tan \frac{\mathrm{x}}{2}}=\frac{\mathrm{r}}{1-\mathrm{r}}=\frac{\sqrt{5}-1}{2}$$$$$$$$\mathbf{soit}:\mathbf{r}=\frac{3-\sqrt{5}}{2}$$$$$$

Commented by a.lgnaoui last updated on 31/Oct/24

Answered by A5T last updated on 01/Nov/24

Commented by A5T last updated on 01/Nov/24

$$y=r...\left(i\right);{(1-r+x)}^2=r^2+{(r+y)}^2=5r^2...\left(ii\right)$$$${(1-r)}^2+r^2={(x+y)}^2+r^2\Rightarrow x=1-2r...\left(iii\right)$$$$\left(iii\right)in\left(ii\right)\Rightarrow {(1-r+1-2r)}^2=5r^2$$$$\Rightarrow {(2-3r)}^2=5r^2\Rightarrow 4r^2-12r+4=0\Rightarrow r=\frac{3-\sqrt{5}}{2}$$

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