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Question-194270

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Question Number 194270 by Tawa11 last updated on 01/Jul/23
Commented by mr W last updated on 02/Jul/23
R:r=3:2
$${R}:{r}=\mathrm{3}:\mathrm{2} \\ $$
Answered by mr W last updated on 02/Jul/23
Commented by mr W last updated on 02/Jul/23
OF=R−r FE=(R−r) cos θ OE=(R−r) sin θ CE=EB=(√(r^2 −(R−r)^2 cos^2 θ)) b=CB=2(√(r^2 −(R−r)^2 cos^2 θ)) a=AB=R−(R−r)sin θ−(√(r^2 −(R−r)^2 cos^2 θ)) c=DC=R+(R−r)sin θ−(√(r^2 −(R−r)^2 cos^2 θ)) a:b:c=2:7:3 ⇒a=2k, b=7k, c=3k R−(R−r)sin θ−(√(r^2 −(R−r)^2 cos^2 θ))=2k ...(u) 2(√(r^2 −(R−r)^2 cos^2 θ))=7k ...(ii) R+(R−r)sin θ−(√(r^2 −(R−r)^2 cos^2 θ))=3k ...(iii) (i): R−(R−r)sin θ=((11k)/2) (iii): R+(R−r)sin θ=((13k)/2) ⇒R=6k ⇒(R−r) sin θ=(k/2) 2(√(r^2 −(R−r)^2 cos^2 θ))=7k 2(√(r^2 −(R−r)^2 +(R−r)^2 sin^2 θ))=7k (√(2Rr−R^2 +(k^2 /4)))=((7k)/2) (√(12kr−36k^2 +(k^2 /4)))=((7k)/2) (√(48kr−143k^2 ))=7k ⇒r=4k ⇒(R/r)=((6k)/(4k))=(3/2)✓ (R−r) sin θ=(k/2) (6k−4k) sin θ=(k/2) sin θ=(1/4) ⇒θ=sin^(−1) (1/4)≈14.48°
$${OF}={R}−{r} \\ $$$${FE}=\left({R}−{r}\right)\:\mathrm{cos}\:\theta \\ $$$${OE}=\left({R}−{r}\right)\:\mathrm{sin}\:\theta \\ $$$${CE}={EB}=\sqrt{{r}^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$${b}={CB}=\mathrm{2}\sqrt{{r}^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$${a}={AB}={R}−\left({R}−{r}\right)\mathrm{sin}\:\theta−\sqrt{{r}^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$${c}={DC}={R}+\left({R}−{r}\right)\mathrm{sin}\:\theta−\sqrt{{r}^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$${a}:{b}:{c}=\mathrm{2}:\mathrm{7}:\mathrm{3}\: \\ $$$$\Rightarrow{a}=\mathrm{2}{k},\:{b}=\mathrm{7}{k},\:{c}=\mathrm{3}{k} \\ $$$${R}−\left({R}−{r}\right)\mathrm{sin}\:\theta−\sqrt{{r}^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}=\mathrm{2}{k}\:\:\:…\left({u}\right) \\ $$$$\mathrm{2}\sqrt{{r}^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}=\mathrm{7}{k}\:\:\:…\left({ii}\right) \\ $$$${R}+\left({R}−{r}\right)\mathrm{sin}\:\theta−\sqrt{{r}^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}=\mathrm{3}{k}\:\:\:…\left({iii}\right) \\ $$$$ \\ $$$$\left({i}\right):\:{R}−\left({R}−{r}\right)\mathrm{sin}\:\theta=\frac{\mathrm{11}{k}}{\mathrm{2}} \\ $$$$\left({iii}\right):\:{R}+\left({R}−{r}\right)\mathrm{sin}\:\theta=\frac{\mathrm{13}{k}}{\mathrm{2}} \\ $$$$\Rightarrow{R}=\mathrm{6}{k} \\ $$$$\Rightarrow\left({R}−{r}\right)\:\mathrm{sin}\:\theta=\frac{{k}}{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{2}\sqrt{{r}^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}=\mathrm{7}{k}\: \\ $$$$\mathrm{2}\sqrt{{r}^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} +\left({R}−{r}\right)^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}=\mathrm{7}{k}\: \\ $$$$\sqrt{\mathrm{2}{Rr}−{R}^{\mathrm{2}} +\frac{{k}^{\mathrm{2}} }{\mathrm{4}}}=\frac{\mathrm{7}{k}}{\mathrm{2}} \\ $$$$\sqrt{\mathrm{12}{kr}−\mathrm{36}{k}^{\mathrm{2}} +\frac{{k}^{\mathrm{2}} }{\mathrm{4}}}=\frac{\mathrm{7}{k}}{\mathrm{2}} \\ $$$$\sqrt{\mathrm{48}{kr}−\mathrm{143}{k}^{\mathrm{2}} }=\mathrm{7}{k} \\ $$$$\Rightarrow{r}=\mathrm{4}{k} \\ $$$$ \\ $$$$\Rightarrow\frac{{R}}{{r}}=\frac{\mathrm{6}{k}}{\mathrm{4}{k}}=\frac{\mathrm{3}}{\mathrm{2}}\checkmark \\ $$$$ \\ $$$$\left({R}−{r}\right)\:\mathrm{sin}\:\theta=\frac{{k}}{\mathrm{2}} \\ $$$$\left(\mathrm{6}{k}−\mathrm{4}{k}\right)\:\mathrm{sin}\:\theta=\frac{{k}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\theta=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{4}}\approx\mathrm{14}.\mathrm{48}° \\ $$
Commented by Tawa11 last updated on 02/Jul/23
Wow, God bless you sir. I really appreciate.
$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

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