Question and Answers Forum

All Questions      Topic List

Mensuration Questions

Previous in All Question      Next in All Question      

Previous in Mensuration      Next in Mensuration      

Question Number 185372 by Mingma last updated on 20/Jan/23

Answered by a.lgnaoui last updated on 21/Jan/23

$$△\mathrm{ODO}\mathrm{Triangle}\mathrm{rectangle}$$$${\mathrm{O}}_1{\mathrm{O}}_2^2={({\mathrm{O}}_1\mathrm{C}+\mathrm{CD})}^2+{\mathrm{O}}_2{\mathrm{D}}^2$$$${\mathrm{O}}_1{\mathrm{O}}_2=1+\mathrm{r}{\mathrm{O}}_1\mathrm{C}=\cos \theta$$$$\mathrm{CD}=\mathrm{rcos}\theta {\mathrm{DO}}_2=\mathrm{r}$$$${(1+\mathrm{r})}^2=\cos {\theta }^2{(1+r)}^2+r^2$$$$1+2r=r^2\cos {}^2\theta +2r{\cos }^2\theta +\cos {}^2\theta$$$$1+2r(1-\cos {}^2\theta )-(r^2+1){\cos }^2\theta =0$$$$r^2-\frac{2r\sin {}^2\theta }{\cos {}^2\theta }-\frac{1}{\cos {}^2\theta }+1=0$$$$r^2-2r{\tan }^2\theta -{\tan }^2\theta =0$$$${(r-{\tan }^2\theta )}^2-{2\tan }^2\theta =0$$$$r=\sqrt{2}\tan \theta +\tan {}^2\theta$$$$$$$$\sin \theta =\frac{r}{1+r}\cos \theta =\sqrt{1-\frac{r^2}{{(1+r)}^2}}$$$$\tan \theta =\frac{r}{\sqrt{1+2r}}$$$$r=\sqrt{2}\left(\frac{r}{\sqrt{1+2r}}\right)+\frac{r^2}{1+2r}$$$$r=\frac{r\sqrt{2(1+2r)}+r^2}{1+2r}$$$$r^2-2r-1=0$$$${(r-1)}^2-2=0$$$$$$$$\mathit{r}=1+\sqrt{2}$$$$$$

Commented by mr W last updated on 21/Jan/23

$$totallywrong!$$$${it}^{\prime }sclearr<1.$$

Commented by a.lgnaoui last updated on 21/Jan/23

Answered by mr W last updated on 21/Jan/23

$${(-\frac{1}{2}+\frac{1}{1}+\frac{2}{r})}^2=2(\frac{1}{2^2}+\frac{1}{1^2}+\frac{2}{r^2})$$$$\Rightarrow r=\frac{8}{9}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com