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Question-52360




Question Number 52360 by ajfour last updated on 06/Jan/19
Commented by ajfour last updated on 06/Jan/19
If θ_(max) = 150° and θ_(min) = 75°  , find  a and b in terms of radius R.
Ifθmax=150°andθmin=75°,findaandbintermsofradiusR.
Answered by mr W last updated on 06/Jan/19
Commented by ajfour last updated on 06/Jan/19
how come the perpendicular  bisector Sir? for certain?!
howcometheperpendicularbisectorSir?forcertain?!
Commented by mr W last updated on 06/Jan/19
let MP=h, AM=MB=d  (d/h)=tan (θ_(max) /2)=m  ⇒d=mh  (d/(h+2R))=tan (θ_(min) /2)=n  ⇒d=n(h+2R)  ⇒mh=n(h+2R)  ⇒h=((2nR)/(m−n))  ⇒d=((2mnR)/(m−n))  a=(√(d^2 +(h+R)^2 ))=(√((((2mnR)/(m−n)))^2 +(((2nR)/(m−n))+R)^2 ))  ⇒a=((√(4m^2 n^2 +(m+n)^2 ))/(m−n))×R  (b/(2d))=((h+R)/a)  b=((2d(h+R))/a)=((2×2mn)/( (√(4m^2 n^2 +(m+n)^2 ))))(((2n)/(m−n))+1)×R  ⇒b=((4mn(m+n))/((m−n)(√(4m^2 n^2 +(m+n)^2 ))))×R  with θ_(max) =150°, θ_(min) =75°  m=tan ((150°)/2)=tan 75°=(√3)+2  n=tan ((75°)/2)  m=((2n)/(1−n^2 ))  mn^2 +2n−m=0  n=(((√(1+m^2 ))−1)/m)=2((√(2−(√3)))−1)+(√3)  m+n=2((√3)+(√(2−(√3))))  m−n=2(2−(√(2−(√3))))  mn=2(√(2+(√3)))−1  ...
letMP=h,AM=MB=ddh=tanθmax2=md=mhdh+2R=tanθmin2=nd=n(h+2R)mh=n(h+2R)h=2nRmnd=2mnRmna=d2+(h+R)2=(2mnRmn)2+(2nRmn+R)2a=4m2n2+(m+n)2mn×Rb2d=h+Rab=2d(h+R)a=2×2mn4m2n2+(m+n)2(2nmn+1)×Rb=4mn(m+n)(mn)4m2n2+(m+n)2×Rwithθmax=150°,θmin=75°m=tan150°2=tan75°=3+2n=tan75°2m=2n1n2mn2+2nm=0n=1+m21m=2(231)+3m+n=2(3+23)mn=2(223)mn=22+31
Commented by mr W last updated on 06/Jan/19
for θ=constant, the locus of point P  is a circle through points A and B.  AB is a chord of this circle. such that  θ_(max)  happens, this circle must tangent  the circle with radius R.
forθ=constant,thelocusofpointPisacirclethroughpointsAandB.ABisachordofthiscircle.suchthatθmaxhappens,thiscirclemusttangentthecirclewithradiusR.
Commented by mr W last updated on 06/Jan/19
Commented by ajfour last updated on 06/Jan/19
fine Sir, let me check with b=0,  in a couple of minutes, please..
fineSir,letmecheckwithb=0,inacoupleofminutes,please..
Commented by mr W last updated on 06/Jan/19
i think i made a mistake. it is correct  that the circle through AB should  tangent the circle with radius R, but  AB is not always centric to the circle  with radius R. but i have assumed  they are centric to each other.
ithinkimadeamistake.itiscorrectthatthecirclethroughABshouldtangentthecirclewithradiusR,butABisnotalwayscentrictothecirclewithradiusR.butihaveassumedtheyarecentrictoeachother.
Commented by mr W last updated on 06/Jan/19
i want to say, my solution is correct,   but there are also other possibilities.
iwanttosay,mysolutioniscorrect,buttherearealsootherpossibilities.
Commented by mr W last updated on 06/Jan/19
since both a and b are to be determined,  it is ok to assume that AB is centric to  the circle and we get the solution as  i did. but if one of a and b is fixed, then it  is wrong to assume that AB is centric  to the circle.
sincebothaandbaretobedetermined,itisoktoassumethatABiscentrictothecircleandwegetthesolutionasidid.butifoneofaandbisfixed,thenitiswrongtoassumethatABiscentrictothecircle.
Commented by ajfour last updated on 06/Jan/19
yes Sir, i tried to agree but my  sketching did not let me agree totally  about the perpendicular bisector  line you drew, not generally.
yesSir,itriedtoagreebutmysketchingdidnotletmeagreetotallyabouttheperpendicularbisectorlineyoudrew,notgenerally.
Commented by mr W last updated on 07/Jan/19
the situation is following:  when a and b are given, we get unique  θ_(max)  and θ_(min) .  but when θ_(max)  and θ_(min)  are given, we  can not get unique a and b. the solution  i gave is a correct solution, but not the  only solution.
thesituationisfollowing:whenaandbaregiven,wegetuniqueθmaxandθmin.butwhenθmaxandθminaregiven,wecannotgetuniqueaandb.thesolutionigaveisacorrectsolution,butnottheonlysolution.
Commented by ajfour last updated on 07/Jan/19
oh yes, Sir.Thank you so much!  dint occur to me, without your words.
ohyes,Sir.Thankyousomuch!dintoccurtome,withoutyourwords.

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