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Question Number 45270 by ajfour last updated on 11/Oct/18

Commented by ajfour last updated on 11/Oct/18

Find 𝛂 in terms of R and r.

FindαintermsofRandr.

Answered by MrW3 last updated on 11/Oct/18

PT=(r/(tan α))=((r cos α)/(sin α))  PA=(R/(sin α))  AT=PA−PT=((R−r cos α)/(sin α))  AC=R+r  AC^2 =AT^2 +TC^2   (R+r)^2 =(((R−r cos α)^2 )/(sin^2  α))+r^2   (R^2 +2Rr)(1−cos^2  α)=R^2 −2Rr cos α+r^2 cos^2  α  R^2 +2Rr−(R^2 +2Rr)cos^2  α=R^2 −2Rr cos α+r^2 cos^2  α  (R+r)^2 cos^2  α−2Rr cos α−2Rr=0  cos α=((2Rr+(√(4R^2 r^2 +8Rr(R+r)^2 )))/(2(R+r)^2 ))  cos α=((Rr+(√(Rr[(R+r)^2 +Rr])))/((R+r)^2 ))  ⇒α=cos^(−1) ((Rr+(√(Rr[(R+r)^2 +Rr])))/((R+r)^2 ))

PT=rtanα=rcosαsinαPA=RsinαAT=PAPT=RrcosαsinαAC=R+rAC2=AT2+TC2(R+r)2=(Rrcosα)2sin2α+r2(R2+2Rr)(1cos2α)=R22Rrcosα+r2cos2αR2+2Rr(R2+2Rr)cos2α=R22Rrcosα+r2cos2α(R+r)2cos2α2Rrcosα2Rr=0cosα=2Rr+4R2r2+8Rr(R+r)22(R+r)2cosα=Rr+Rr[(R+r)2+Rr](R+r)2α=cos1Rr+Rr[(R+r)2+Rr](R+r)2

Commented by ajfour last updated on 11/Oct/18

Thank you Sir !

ThankyouSir!

Answered by ajfour last updated on 11/Oct/18

PM=((Rcos α)/(sin α))=(r/(sin α))+(√((R+r)^2 −R^2 ))  AP = (R/(sin α)) = ((rcos α)/(sin α))+(√((R+r)^2 −r^2 ))  ⇒ ((Rcos α−r)/(R−rcos α)) = ((√(r(2R+r)))/(√(R(2r+R)))) = (a/b) (let)  ⇒ bRcos α−br = aR−arcos α     cos α = ((aR+br)/(ar+bR))    𝛂 = cos^(−1) (((R(√(r(2R+r)))+r(√(R(2r+R))))/(r(√(r(2R+r)))+R(√(R(2r+R)))))) .

PM=Rcosαsinα=rsinα+(R+r)2R2AP=Rsinα=rcosαsinα+(R+r)2r2RcosαrRrcosα=r(2R+r)R(2r+R)=ab(let)bRcosαbr=aRarcosαcosα=aR+brar+bRα=cos1(Rr(2R+r)+rR(2r+R)rr(2R+r)+RR(2r+R)).

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