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Question Number 213173 by ajfour last updated on 31/Oct/24
Commented by ajfour last updated on 31/Oct/24
Find r
$${Find}\:{r} \\ $$
Answered by ajfour last updated on 31/Oct/24
Commented by ajfour last updated on 01/Nov/24
φ=((90°−2θ)/2)=45°−θ rtan φ=r(((1−tan θ)/(1+tan θ)))=1−2r ⇒ tan θ=((r−(1−2r))/(r+(1−2r)))=((3r−1)/(1−r)) (x+r)^2 +1=(x+1−r)^2 ⇒ 2rx=2x−2r−2rx x=(r/(1−2r)) Further (r/x)=tan θ=((3r−1)/(1−r)) ⇒ x=((r(1−r))/(3r−1))=(r/(1−2r)) ⇒ (1−r)(1−2r)=3r−1 ⇒ 2r^2 −6r+2=0 ⇒ r^2 −3r+1=0 r=(3/2)−(√((9/4)−1)) r=((3−(√5))/2)
$$\phi=\frac{\mathrm{90}°−\mathrm{2}\theta}{\mathrm{2}}=\mathrm{45}°−\theta \\ $$$${r}\mathrm{tan}\:\phi={r}\left(\frac{\mathrm{1}−\mathrm{tan}\:\theta}{\mathrm{1}+\mathrm{tan}\:\theta}\right)=\mathrm{1}−\mathrm{2}{r} \\ $$$$\Rightarrow\:\:\mathrm{tan}\:\theta=\frac{{r}−\left(\mathrm{1}−\mathrm{2}{r}\right)}{{r}+\left(\mathrm{1}−\mathrm{2}{r}\right)}=\frac{\mathrm{3}{r}−\mathrm{1}}{\mathrm{1}−{r}} \\ $$$$\left({x}+{r}\right)^{\mathrm{2}} +\mathrm{1}=\left({x}+\mathrm{1}−{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{2}{rx}=\mathrm{2}{x}−\mathrm{2}{r}−\mathrm{2}{rx} \\ $$$${x}=\frac{{r}}{\mathrm{1}−\mathrm{2}{r}} \\ $$$${Further} \\ $$$$\frac{{r}}{{x}}=\mathrm{tan}\:\theta=\frac{\mathrm{3}{r}−\mathrm{1}}{\mathrm{1}−{r}} \\ $$$$\Rightarrow\:\:{x}=\frac{{r}\left(\mathrm{1}−{r}\right)}{\mathrm{3}{r}−\mathrm{1}}=\frac{{r}}{\mathrm{1}−\mathrm{2}{r}} \\ $$$$\Rightarrow\:\:\:\left(\mathrm{1}−{r}\right)\left(\mathrm{1}−\mathrm{2}{r}\right)=\mathrm{3}{r}−\mathrm{1} \\ $$$$\Rightarrow\:\:\:\mathrm{2}{r}^{\mathrm{2}} −\mathrm{6}{r}+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\:\:{r}^{\mathrm{2}} −\mathrm{3}{r}+\mathrm{1}=\mathrm{0} \\ $$$${r}=\frac{\mathrm{3}}{\mathrm{2}}−\sqrt{\frac{\mathrm{9}}{\mathrm{4}}−\mathrm{1}} \\ $$$${r}=\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$
Commented by Frix last updated on 31/Oct/24
Your answer is right.
$$\mathrm{Your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{right}. \\ $$
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 2θ comes out equal to 2.
$$\mathrm{tan}\:\mathrm{2}\theta\:\:{comes}\:{out}\:{equal}\:{to}\:\mathrm{2}. \\ $$
Answered by a.lgnaoui last updated on 31/Oct/24
tan x=(r/(2r))=(1/2)=((2t)/(1−t^2 )) t=tan (x/2) t^2 +4t−1=0 t=((−4±2(√5))/2)=(√5)−2 tan (((90−x)/2))=(r/(1−r)) ((tan 45−tan (x/2))/(1+tan (x/2)))=(r/(1−r))=(((√5) −1)/2) soit: r=((3−(√5))/2)
$$\mathrm{tan}\:\mathrm{x}=\frac{\mathrm{r}}{\mathrm{2r}}=\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{2t}}{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }\:\:\:\:\mathrm{t}=\mathrm{tan}\:\frac{\mathrm{x}}{\mathrm{2}} \\ $$$$\mathrm{t}^{\mathrm{2}} +\mathrm{4t}−\mathrm{1}=\mathrm{0}\:\:\:\:\:\mathrm{t}=\frac{−\mathrm{4}\pm\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{2}}=\sqrt{\mathrm{5}}−\mathrm{2} \\ $$$$ \\ $$$$\mathrm{tan}\:\left(\frac{\mathrm{90}−\mathrm{x}}{\mathrm{2}}\right)=\frac{\mathrm{r}}{\mathrm{1}−\mathrm{r}} \\ $$$$\:\frac{\mathrm{tan}\:\mathrm{45}−\mathrm{tan}\:\frac{\mathrm{x}}{\mathrm{2}}}{\mathrm{1}+\mathrm{tan}\:\frac{\mathrm{x}}{\mathrm{2}}}=\frac{\mathrm{r}}{\mathrm{1}−\mathrm{r}}=\frac{\sqrt{\mathrm{5}}\:−\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{soit}}:\:\:\:\:\:\boldsymbol{\mathrm{r}}=\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\:\:\: \\ $$
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...(i);(1−r+x)^2 =r^2 +(r+y)^2 =5r^2 ...(ii) (1−r)^2 +r^2 =(x+y)^2 +r^2 ⇒x=1−2r...(iii) (iii) in (ii)⇒(1−r+1−2r)^2 =5r^2 ⇒(2−3r)^2 =5r^2 ⇒4r^2 −12r+4=0⇒r=((3−(√5))/2)
$${y}={r}…\left({i}\right);\left(\mathrm{1}−{r}+{x}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} +\left({r}+{y}\right)^{\mathrm{2}} =\mathrm{5}{r}^{\mathrm{2}} …\left({ii}\right) \\ $$$$\left(\mathrm{1}−{r}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} =\left({x}+{y}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} \Rightarrow{x}=\mathrm{1}−\mathrm{2}{r}…\left({iii}\right) \\ $$$$\left({iii}\right)\:{in}\:\left({ii}\right)\Rightarrow\left(\mathrm{1}−{r}+\mathrm{1}−\mathrm{2}{r}\right)^{\mathrm{2}} =\mathrm{5}{r}^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{2}−\mathrm{3}{r}\right)^{\mathrm{2}} =\mathrm{5}{r}^{\mathrm{2}} \Rightarrow\mathrm{4}{r}^{\mathrm{2}} −\mathrm{12}{r}+\mathrm{4}=\mathrm{0}\Rightarrow{r}=\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

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