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Question Number 162100 by amin96 last updated on 26/Dec/21

Commented by mr W last updated on 27/Dec/21

tan 𝛂=((2(Rβˆ’r))/(R+r))

$$\boldsymbol{\mathrm{tan}}\:\boldsymbol{\alpha}=\frac{\mathrm{2}\left(\boldsymbol{{R}}βˆ’\boldsymbol{{r}}\right)}{\boldsymbol{{R}}+\boldsymbol{{r}}} \\ $$

Answered by mr W last updated on 27/Dec/21

Commented by mr W last updated on 27/Dec/21

2p+2r=2R β‡’p=Rβˆ’r BC=Rβˆ’r CD=Rβˆ’p=Rβˆ’(Rβˆ’r)=r BE=r+q CE=Rβˆ’q DE=p+q=Rβˆ’r+q ((BC^2 +CE^2 βˆ’BE^2 )/(2Γ—BCΓ—CE))=βˆ’((CD^2 +CE^2 βˆ’DE^2 )/(2Γ—CDΓ—CE)) (((Rβˆ’r)^2 +(Rβˆ’q)^2 βˆ’(r+q)^2 )/(Rβˆ’r))=βˆ’((r^2 +(Rβˆ’q)^2 βˆ’(Rβˆ’r+q)^2 )/r) ((R^2 βˆ’Rrβˆ’(R+r)q)/(Rβˆ’r))=βˆ’((Rrβˆ’(2Rβˆ’r)q)/r) (R^2 βˆ’Rr+r^2 )q=(Rβˆ’r)Rr β‡’q=(((Rβˆ’r)Rr)/(R^2 βˆ’Rr+r^2 )) BC+CE+EB=(Rβˆ’r)+(Rβˆ’q)+(r+q)=2R βˆ’BC+CE+EB=βˆ’(Rβˆ’r)+(Rβˆ’q)+(r+q)=2r BCβˆ’CE+EB=(Rβˆ’r)βˆ’(Rβˆ’q)+(r+q)=2q BC+CEβˆ’EB=(Rβˆ’r)+(Rβˆ’q)βˆ’(r+q)=2(Rβˆ’rβˆ’q) Ξ”_(BCE) =((√(2RΓ—2rΓ—2qΓ—2(Rβˆ’rβˆ’q)))/4)=(√(Rrq(Rβˆ’rβˆ’q))) Ξ”_(BCE) =((BCΓ—h)/2) β‡’h=((2(√(Rrq(Rβˆ’rβˆ’q))))/(Rβˆ’r))=((2(Rβˆ’r)Rr)/(R^2 βˆ’Rr+r^2 ))=2q (√(BE^2 βˆ’h^2 ))=(√((r+q)^2 βˆ’4q^2 ))=(√((r+3q)(rβˆ’q))) tan Ξ±=(h/(AB+(√(BE^2 βˆ’h^2 )))) =((2q)/(r+(√((r+3q)(rβˆ’q))))) =((2q)/(r+(√((r+((3(Rβˆ’r)Rr)/(R^2 βˆ’Rr+r^2 )))(rβˆ’(((Rβˆ’r)Rr)/(R^2 βˆ’Rr+r^2 ))))))) =(2/(r+((r^2 (2Rβˆ’r))/(R^2 βˆ’Rr+r^2 ))))Γ—(((Rβˆ’r)Rr)/(R^2 βˆ’Rr+r^2 )) =((2(Rβˆ’r))/(R+r)) check for special case R=2r: tan Ξ±=((2(2rβˆ’r))/(2r+r))=(2/3) (r+q)^2 =r^2 +(2rβˆ’q)^2 q=((2r)/3) tan Ξ±=((2rβˆ’q)/(2r))=((2rβˆ’((2r)/3))/(2r))=(2/3) βœ“

$$\mathrm{2}{p}+\mathrm{2}{r}=\mathrm{2}{R}\:\Rightarrow{p}={R}βˆ’{r} \\ $$$${BC}={R}βˆ’{r} \\ $$$${CD}={R}βˆ’{p}={R}βˆ’\left({R}βˆ’{r}\right)={r} \\ $$$${BE}={r}+{q} \\ $$$${CE}={R}βˆ’{q} \\ $$$${DE}={p}+{q}={R}βˆ’{r}+{q} \\ $$$$\frac{{BC}^{\mathrm{2}} +{CE}^{\mathrm{2}} βˆ’{BE}^{\mathrm{2}} }{\mathrm{2}Γ—{BC}Γ—{CE}}=βˆ’\frac{{CD}^{\mathrm{2}} +{CE}^{\mathrm{2}} βˆ’{DE}^{\mathrm{2}} }{\mathrm{2}Γ—{CD}Γ—{CE}} \\ $$$$\frac{\left({R}βˆ’{r}\right)^{\mathrm{2}} +\left({R}βˆ’{q}\right)^{\mathrm{2}} βˆ’\left({r}+{q}\right)^{\mathrm{2}} }{{R}βˆ’{r}}=βˆ’\frac{{r}^{\mathrm{2}} +\left({R}βˆ’{q}\right)^{\mathrm{2}} βˆ’\left({R}βˆ’{r}+{q}\right)^{\mathrm{2}} }{{r}} \\ $$$$\frac{{R}^{\mathrm{2}} βˆ’{Rr}βˆ’\left({R}+{r}\right){q}}{{R}βˆ’{r}}=βˆ’\frac{{Rr}βˆ’\left(\mathrm{2}{R}βˆ’{r}\right){q}}{{r}} \\ $$$$\left({R}^{\mathrm{2}} βˆ’{Rr}+{r}^{\mathrm{2}} \right){q}=\left({R}βˆ’{r}\right){Rr} \\ $$$$\Rightarrow{q}=\frac{\left({R}βˆ’{r}\right){Rr}}{{R}^{\mathrm{2}} βˆ’{Rr}+{r}^{\mathrm{2}} } \\ $$$${BC}+{CE}+{EB}=\left({R}βˆ’{r}\right)+\left({R}βˆ’{q}\right)+\left({r}+{q}\right)=\mathrm{2}{R} \\ $$$$βˆ’{BC}+{CE}+{EB}=βˆ’\left({R}βˆ’{r}\right)+\left({R}βˆ’{q}\right)+\left({r}+{q}\right)=\mathrm{2}{r} \\ $$$${BC}βˆ’{CE}+{EB}=\left({R}βˆ’{r}\right)βˆ’\left({R}βˆ’{q}\right)+\left({r}+{q}\right)=\mathrm{2}{q} \\ $$$${BC}+{CE}βˆ’{EB}=\left({R}βˆ’{r}\right)+\left({R}βˆ’{q}\right)βˆ’\left({r}+{q}\right)=\mathrm{2}\left({R}βˆ’{r}βˆ’{q}\right) \\ $$$$\Delta_{{BCE}} =\frac{\sqrt{\mathrm{2}{R}Γ—\mathrm{2}{r}Γ—\mathrm{2}{q}Γ—\mathrm{2}\left({R}βˆ’{r}βˆ’{q}\right)}}{\mathrm{4}}=\sqrt{{Rrq}\left({R}βˆ’{r}βˆ’{q}\right)} \\ $$$$\Delta_{{BCE}} =\frac{{BC}Γ—{h}}{\mathrm{2}} \\ $$$$\Rightarrow{h}=\frac{\mathrm{2}\sqrt{{Rrq}\left({R}βˆ’{r}βˆ’{q}\right)}}{{R}βˆ’{r}}=\frac{\mathrm{2}\left({R}βˆ’{r}\right){Rr}}{{R}^{\mathrm{2}} βˆ’{Rr}+{r}^{\mathrm{2}} }=\mathrm{2}{q} \\ $$$$\sqrt{{BE}^{\mathrm{2}} βˆ’{h}^{\mathrm{2}} }=\sqrt{\left({r}+{q}\right)^{\mathrm{2}} βˆ’\mathrm{4}{q}^{\mathrm{2}} }=\sqrt{\left({r}+\mathrm{3}{q}\right)\left({r}βˆ’{q}\right)} \\ $$$$\mathrm{tan}\:\alpha=\frac{{h}}{{AB}+\sqrt{{BE}^{\mathrm{2}} βˆ’{h}^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{2}{q}}{{r}+\sqrt{\left({r}+\mathrm{3}{q}\right)\left({r}βˆ’{q}\right)}} \\ $$$$=\frac{\mathrm{2}{q}}{{r}+\sqrt{\left({r}+\frac{\mathrm{3}\left({R}βˆ’{r}\right){Rr}}{{R}^{\mathrm{2}} βˆ’{Rr}+{r}^{\mathrm{2}} }\right)\left({r}βˆ’\frac{\left({R}βˆ’{r}\right){Rr}}{{R}^{\mathrm{2}} βˆ’{Rr}+{r}^{\mathrm{2}} }\right)}} \\ $$$$=\frac{\mathrm{2}}{{r}+\frac{{r}^{\mathrm{2}} \left(\mathrm{2}{R}βˆ’{r}\right)}{{R}^{\mathrm{2}} βˆ’{Rr}+{r}^{\mathrm{2}} }}Γ—\frac{\left({R}βˆ’{r}\right){Rr}}{{R}^{\mathrm{2}} βˆ’{Rr}+{r}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}\left({R}βˆ’{r}\right)}{{R}+{r}} \\ $$$${check}\:{for}\:{special}\:{case}\:{R}=\mathrm{2}{r}: \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{2}\left(\mathrm{2}{r}βˆ’{r}\right)}{\mathrm{2}{r}+{r}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$ \\ $$$$\left({r}+{q}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} +\left(\mathrm{2}{r}βˆ’{q}\right)^{\mathrm{2}} \\ $$$${q}=\frac{\mathrm{2}{r}}{\mathrm{3}} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{2}{r}βˆ’{q}}{\mathrm{2}{r}}=\frac{\mathrm{2}{r}βˆ’\frac{\mathrm{2}{r}}{\mathrm{3}}}{\mathrm{2}{r}}=\frac{\mathrm{2}}{\mathrm{3}}\:\checkmark \\ $$

Commented by Tawa11 last updated on 27/Dec/21

Great sir. Am always gaining from your geometry solution.

$$\mathrm{Great}\:\mathrm{sir}.\:\mathrm{Am}\:\mathrm{always}\:\mathrm{gaining}\:\mathrm{from}\:\mathrm{your}\:\mathrm{geometry}\:\mathrm{solution}. \\ $$

Commented by Tawa11 last updated on 27/Dec/21

God bless you more sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{more}\:\mathrm{sir} \\ $$

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