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Question-198479

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Question Number 198479 by ajfour last updated on 20/Oct/23
Commented by ajfour last updated on 20/Oct/23
Find r if α be known. Outer circle has unit radius.
$${Find}\:{r}\:{if}\:\alpha\:{be}\:{known}.\:{Outer}\:{circle} \\ $$$${has}\:{unit}\:{radius}. \\ $$
Commented by dangduomg last updated on 21/Oct/23
does the inner circle havw to be tangential to outer circle?
$${does}\:{the}\:{inner}\:{circle}\:{havw}\:{to}\:{be}\:{tangential}\:{to}\:{outer}\:{circle}? \\ $$
Answered by mr W last updated on 21/Oct/23
((r/(tan α))−1)^2 +r^2 =(1−r)^2 ⇒r=2(1−tan α)tan α
$$\left(\frac{{r}}{\mathrm{tan}\:\alpha}−\mathrm{1}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} =\left(\mathrm{1}−{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{r}=\mathrm{2}\left(\mathrm{1}−\mathrm{tan}\:\alpha\right)\mathrm{tan}\:\alpha \\ $$
Commented by mr W last updated on 21/Oct/23
Commented by ajfour last updated on 21/Oct/23
yes sir, short correct and beautiful solution.
Commented by Hridiana last updated on 21/Oct/23
r=cos((θ/2))+(√θ)
$${r}={cos}\left(\frac{\theta}{\mathrm{2}}\right)+\sqrt{\theta} \\ $$
Commented by Hridiana last updated on 21/Oct/23
theres nothing much like α
$${theres}\:{nothing}\:{much}\:{like}\:\alpha \\ $$
Answered by a.lgnaoui last updated on 21/Oct/23
△MIF sin 𝛂=((MI)/(IF))=((BC)/(BF)) (i) △OMI ∡MOI=𝛉 sin 𝛉=((MI)/(OI))=(r/(1−r)) ⇒ sin 𝛉=(r/(1−r)) cos 𝛉=((√(1−2r))/(1−r)) ⇒MF=1+(√(1−2r)) BC=1×sin 2𝛂 CF=1+cos 2𝛂 tan 𝛂=((BC)/(CF))=((MI)/(IF))= (r/(1+(√(1−2r)))) posons t=tan 𝛂 t(1+(√(1−2r)) )=r t (√(1−2r)) =r−t t^2 (1−2r)=r^2 +t^2 −2rt r=2t(1−t)
$$\bigtriangleup\mathrm{MIF}\:\:\:\mathrm{sin}\:\boldsymbol{\alpha}=\frac{\boldsymbol{\mathrm{MI}}}{\mathrm{I}\boldsymbol{\mathrm{F}}}=\frac{\boldsymbol{\mathrm{BC}}}{\boldsymbol{\mathrm{BF}}}\:\:\:\:\left(\boldsymbol{\mathrm{i}}\right) \\ $$$$\bigtriangleup\mathrm{OMI}\:\:\:\measuredangle\boldsymbol{\mathrm{MOI}}=\boldsymbol{\theta}\:\:\:\mathrm{sin}\:\boldsymbol{\theta}=\frac{\boldsymbol{\mathrm{MI}}}{\boldsymbol{\mathrm{OI}}}=\frac{\boldsymbol{\mathrm{r}}}{\mathrm{1}−\boldsymbol{\mathrm{r}}} \\ $$$$\Rightarrow\:\mathrm{sin}\:\boldsymbol{\theta}=\frac{\boldsymbol{\mathrm{r}}}{\mathrm{1}−\boldsymbol{\mathrm{r}}}\:\:\:\mathrm{cos}\:\boldsymbol{\theta}=\frac{\sqrt{\mathrm{1}−\mathrm{2}\boldsymbol{\mathrm{r}}}}{\mathrm{1}−\boldsymbol{\mathrm{r}}} \\ $$$$\Rightarrow\boldsymbol{\mathrm{MF}}=\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{2}\boldsymbol{\mathrm{r}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\boldsymbol{\mathrm{BC}}=\mathrm{1}×\mathrm{sin}\:\:\mathrm{2}\boldsymbol{\alpha}\:\:\:\:\:\boldsymbol{\mathrm{CF}}=\mathrm{1}+\mathrm{cos}\:\mathrm{2}\boldsymbol{\alpha} \\ $$$$\mathrm{tan}\:\boldsymbol{\alpha}=\frac{\boldsymbol{\mathrm{BC}}}{\boldsymbol{\mathrm{CF}}}=\frac{\boldsymbol{\mathrm{MI}}}{\boldsymbol{\mathrm{IF}}}=\:\frac{\boldsymbol{\mathrm{r}}}{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{2}\boldsymbol{\mathrm{r}}}} \\ $$$$\:\:\:\:\mathrm{posons}\:\:\:\boldsymbol{\mathrm{t}}=\mathrm{tan}\:\boldsymbol{\alpha} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{t}}\left(\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{2}\boldsymbol{\mathrm{r}}}\:\right)=\boldsymbol{\mathrm{r}} \\ $$$$\:\:\boldsymbol{\mathrm{t}}\:\:\sqrt{\mathrm{1}−\mathrm{2}\boldsymbol{\mathrm{r}}}\:\:=\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{t}}\:\:\: \\ $$$$\:\:\boldsymbol{\mathrm{t}}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{2}\boldsymbol{\mathrm{r}}\right)=\boldsymbol{\mathrm{r}}^{\mathrm{2}} +\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{rt}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{r}}=\mathrm{2}\boldsymbol{\mathrm{t}}\left(\mathrm{1}−\boldsymbol{\mathrm{t}}\right) \\ $$$$\: \\ $$
Commented by a.lgnaoui last updated on 21/Oct/23

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