Question-185372

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

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

△ODO Triangle rectangle O_1 O_2 ^2 =(O_1 C+CD)^2 +O_2 D^2 O_1 O_2 =1+r O_1 C=cos θ CD=rcos θ DO_2 =r (1+r)^2 =cosθ^2 (1+r)^2 +r^2 1+2r=r^2 cos^2 θ+2rcos^2 θ+cos^2 θ 1+2r(1−cos^2 θ)−(r^2 +1)cos^2 θ=0 r^2 −((2rsin^2 θ)/(cos^2 θ))−(1/(cos^2 θ))+1=0 r^2 −2rtan^2 θ−tan^2 θ=0 (r−tan^2 θ)^2 −2tan^2 θ =0 r=(√2) tanθ+tan^2 θ sin θ=(r/(1+r)) cos θ=(√(1−(r^2 /((1+r)^2 )))) tan θ=(r/( (√(1+2r)))) r=(√2) ((r/( (√(1+2r)))))+(r^2 /(1+2r)) r=((r(√(2(1+2r))) +r^2 )/(1+2r)) r^2 −2r−1=0 (r−1)^2 −2=0 r=1+(√2)

$$\bigtriangleup\mathrm{ODO}\:\:\:\mathrm{Triangle}\:\mathrm{rectangle} \\ $$$$\mathrm{O}_{\mathrm{1}} \mathrm{O}_{\mathrm{2}} ^{\mathrm{2}} =\left(\mathrm{O}_{\mathrm{1}} \mathrm{C}+\mathrm{CD}\right)^{\mathrm{2}} +\mathrm{O}_{\mathrm{2}} \mathrm{D}^{\mathrm{2}} \\ $$$$\mathrm{O}_{\mathrm{1}} \mathrm{O}_{\mathrm{2}} =\mathrm{1}+\mathrm{r}\:\:\:\:\:\mathrm{O}_{\mathrm{1}} \mathrm{C}=\mathrm{cos}\:\theta \\ $$$$\mathrm{CD}=\mathrm{rcos}\:\theta\:\:\:\:\:\:\mathrm{DO}_{\mathrm{2}} =\mathrm{r} \\ $$$$\left(\mathrm{1}+\mathrm{r}\right)^{\mathrm{2}} =\mathrm{cos}\theta^{\mathrm{2}} \left(\mathrm{1}+{r}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} \\ $$$$\mathrm{1}+\mathrm{2}{r}={r}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta+\mathrm{2}{r}\mathrm{cos}^{\mathrm{2}} \theta+\mathrm{cos}\:^{\mathrm{2}} \theta \\ $$$$\mathrm{1}+\mathrm{2}{r}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \theta\right)−\left({r}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{cos}^{\mathrm{2}} \theta=\mathrm{0} \\ $$$$\:\:{r}^{\mathrm{2}} −\frac{\mathrm{2}{r}\mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{cos}\:^{\mathrm{2}} \theta}−\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} \theta}+\mathrm{1}=\mathrm{0} \\ $$$${r}^{\mathrm{2}} −\mathrm{2}{r}\mathrm{tan}^{\mathrm{2}} \theta−\mathrm{tan}^{\mathrm{2}} \theta=\mathrm{0}\: \\ $$$$\left({r}−\mathrm{tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} −\mathrm{2tan}^{\mathrm{2}} \theta\:=\mathrm{0} \\ $$$${r}=\sqrt{\mathrm{2}}\:\mathrm{tan}\theta+\mathrm{tan}\:^{\mathrm{2}} \theta\: \\ $$$$\: \\ $$$$\mathrm{sin}\:\theta=\frac{{r}}{\mathrm{1}+{r}}\:\:\:\mathrm{cos}\:\theta=\sqrt{\mathrm{1}−\frac{{r}^{\mathrm{2}} }{\left(\mathrm{1}+{r}\right)^{\mathrm{2}} }} \\ $$$$\mathrm{tan}\:\theta=\frac{{r}}{\:\sqrt{\mathrm{1}+\mathrm{2}{r}}} \\ $$$${r}=\sqrt{\mathrm{2}}\:\left(\frac{{r}}{\:\sqrt{\mathrm{1}+\mathrm{2}{r}}}\right)+\frac{{r}^{\mathrm{2}} }{\mathrm{1}+\mathrm{2}{r}} \\ $$$${r}=\frac{{r}\sqrt{\mathrm{2}\left(\mathrm{1}+\mathrm{2}{r}\right)}\:\:+{r}^{\mathrm{2}} }{\mathrm{1}+\mathrm{2}{r}} \\ $$$${r}^{\mathrm{2}} −\mathrm{2}{r}−\mathrm{1}=\mathrm{0} \\ $$$$\left({r}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}=\mathrm{0} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{{r}}=\mathrm{1}+\sqrt{\mathrm{2}} \\ $$$$ \\ $$

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

$${totally}\:{wrong}! \\ $$$${it}'{s}\:{clear}\:{r}<\mathrm{1}. \\ $$

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

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

(−(1/2)+(1/1)+(2/r))^2 =2((1/2^2 )+(1/1^2 )+(2/r^2 )) ⇒r=(8/9)

$$\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{2}}{{r}}\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{2}}{{r}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow{r}=\frac{\mathrm{8}}{\mathrm{9}} \\ $$

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