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Question-161367

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Question Number 161367 by amin96 last updated on 17/Dec/21
Commented by 1549442205PVT last updated on 17/Dec/21
I think that the condition of the probem isn′t clear the radius of the arc which limits small circle isn′t given yet
$${I}\:{think}\:{that}\:{the}\:{condition}\:{of}\:{the}\:{probem}\:{isn}'{t}\:\:{clear} \\ $$$${the}\:{radius}\:{of}\:{the}\:{arc}\:{which}\:{limits}\:{small} \\ $$$${circle}\:{isn}'{t}\:{given}\:{yet} \\ $$
Answered by mr W last updated on 17/Dec/21
Commented by mr W last updated on 17/Dec/21
BC=R−r BE=(√((R−r)^2 −r^2 ))=(√(R^2 −2Rr)) OB=R OE=R+(√(R^2 −2Rr)) OC=(√(r^2 +(R+(√(R^2 −2Rr)))^2 ))=(√(2R^2 −2Rr+r^2 +2R(√(R^2 −2Rr)))) cos α=((OE)/(OC)), sin α=((EC)/(OC)) β=(π/3)−α AC=R+r AC^2 =OA^2 +OC^2 −2×OA×OC×cos ((π/3)−α) AC^2 =OA^2 +OC^2 −OA×OC(cos α+(√3) sin α) AC^2 =OA^2 +OC^2 −OA(OE+(√3) EC) (R+r)^2 =R^2 +2R^2 −2Rr+r^2 +2R(√(R^2 −2Rr))−R(R+(√(R^2 −2Rr))+(√3)r) (√(R^2 −2Rr))=(4+(√3))r−R (4+(√3))^2 r=2(3+(√3))R ⇒(r/R)=((2(3+(√3)))/((4+(√3))^2 ))=((2(33−5(√3)))/(169))≈0.288044
$${BC}={R}−{r} \\ $$$${BE}=\sqrt{\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }=\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rr}} \\ $$$${OB}={R} \\ $$$${OE}={R}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rr}} \\ $$$${OC}=\sqrt{{r}^{\mathrm{2}} +\left({R}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rr}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{2}{R}^{\mathrm{2}} −\mathrm{2}{Rr}+{r}^{\mathrm{2}} +\mathrm{2}{R}\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rr}}} \\ $$$$\mathrm{cos}\:\alpha=\frac{{OE}}{{OC}},\:\mathrm{sin}\:\alpha=\frac{{EC}}{{OC}} \\ $$$$\beta=\frac{\pi}{\mathrm{3}}−\alpha \\ $$$${AC}={R}+{r} \\ $$$${AC}^{\mathrm{2}} ={OA}^{\mathrm{2}} +{OC}^{\mathrm{2}} −\mathrm{2}×{OA}×{OC}×\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}−\alpha\right) \\ $$$${AC}^{\mathrm{2}} ={OA}^{\mathrm{2}} +{OC}^{\mathrm{2}} −{OA}×{OC}\left(\mathrm{cos}\:\alpha+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\alpha\right) \\ $$$${AC}^{\mathrm{2}} ={OA}^{\mathrm{2}} +{OC}^{\mathrm{2}} −{OA}\left({OE}+\sqrt{\mathrm{3}}\:{EC}\right) \\ $$$$\left({R}+{r}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} +\mathrm{2}{R}^{\mathrm{2}} −\mathrm{2}{Rr}+{r}^{\mathrm{2}} +\mathrm{2}{R}\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rr}}−{R}\left({R}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rr}}+\sqrt{\mathrm{3}}{r}\right) \\ $$$$\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rr}}=\left(\mathrm{4}+\sqrt{\mathrm{3}}\right){r}−{R} \\ $$$$\left(\mathrm{4}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} {r}=\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{3}}\right){R} \\ $$$$\Rightarrow\frac{{r}}{{R}}=\frac{\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{3}}\right)}{\left(\mathrm{4}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }=\frac{\mathrm{2}\left(\mathrm{33}−\mathrm{5}\sqrt{\mathrm{3}}\right)}{\mathrm{169}}\approx\mathrm{0}.\mathrm{288044} \\ $$
Commented by Tawa11 last updated on 18/Dec/21
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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