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Find-the-side-lengths-of-a-triangle-if-side-lengths-are-consecutive-integers-and-one-of-whose-angles-is-twice-as-large-as-another-

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Question Number 18968 by chux last updated on 02/Aug/17
Find the side lengths of a triangle if side lengths are consecutive integers,and one of whose angles is twice as large as another.
Findthesidelengthsofatriangleifsidelengthsareconsecutiveintegers,andoneofwhoseanglesistwiceaslargeasanother.
Commented by chux last updated on 02/Aug/17
please help
pleasehelp
Answered by mrW1 last updated on 03/Aug/17
a=n−1 b=n c=n+1 case 1: A=α C=2α B=180−3α ((sin A)/a)=((sin B)/b)=((sin C)/c) ((sin α)/(n−1))=((sin (3α))/n)=((sin (2α))/(n+1)) ((sin (2α))/(sin α))=2cos α=((n−1)/(n+1)) ((sin (3α))/(sin α))=((3 sin α−4 sin^3 α)/(sin α))=3−4sin^2 α=(2cos α)^2 −1=(n/(n−1)) (((n+1)/(n−1)))^2 −1=(n/(n−1)) (((n+1)/(n−1))+1)(((n+1)/(n−1))−1)=(n/(n−1)) (4/(n−1))=1 ⇒n=5 ⇒sides of triangle =4,5,6 case 2: ((sin α)/(n−1))=((sin (2α))/n)=((sin (3α))/(n+1)) ((sin (2α))/(sin α))=2cos α=(n/(n−1)) ((sin (3α))/(sin α))=(2cos α)^2 −1=((n+1)/(n−1)) ((n/(n−1))−1)((n/(n−1))+1)=((n+1)/(n+1)) ((2n−1)/(n−1))=n+1 ⇒n=2 (not suitable, it′s a straight) case 3: ((sin (3α))/(n−1))=((sin α)/n)=((sin (2α))/(n+1)) ((sin (2α))/(sin α))=2cos α=((n+1)/n) ((sin (3α))/(sin α))=(2cos α)^2 −1=((n−1)/n) (((n+1)/n)−1)(((n+1)/n)+1)=((n−1)/n) ⇒n=−2 <0 (not suitable) case 4: ((sin (3α))/(n−1))=((sin (2α))/n)=((sin α)/(n+1)) ((sin (2α))/(sin α))=2cos α=(n/(n+1)) ((sin (3α))/(sin α))=(2cos α)^2 −1=((n−1)/(n+1)) ((n/(n+1))−1)((n/(n+1))+1)=((n−1)/(n+1)) ⇒n=−2<0 (not suitable) case 5: ((sin (2α))/(n−1))=((sin α)/n)=((sin (3α))/(n+1)) ((sin (2α))/(sin α))=2cos α=((n−1)/n) ((sin (3α))/(sin α))=(2cos α)^2 −1=((n+1)/n) (((n−1)/n)+1)(((n−1)/n)−1)=((n+1)/n) n=−2<0 (not suitable) case 6: ((sin (2α))/(n−1))=((sin (3α))/n)=((sin α)/(n+1)) ((sin (2α))/(sin α))=2cos α=((n−1)/(n+1)) ((sin (3α))/(sin α))=(2cos α)^2 −1=(n/(n+1)) (((n−1)/(n+1))+1)(((n−1)/(n+1))−1)=(n/(n+1)) ((−4)/(n+1))=1 ⇒n=−5<0 (not suitable) ⇒the only solution is: 4,5,6
a=n−1b=nc=n+1case1:A=αC=2αB=180−3αsinAa=sinBb=sinCcsinαn−1=sin(3α)n=sin(2α)n+1sin(2α)sinα=2cosα=n−1n+1sin(3α)sinα=3sinα−4sin3αsinα=3−4sin2α=(2cosα)2−1=nn−1(n+1n−1)2−1=nn−1(n+1n−1+1)(n+1n−1−1)=nn−14n−1=1⇒n=5⇒sidesoftriangle=4,5,6case2:sinαn−1=sin(2α)n=sin(3α)n+1sin(2α)sinα=2cosα=nn−1sin(3α)sinα=(2cosα)2−1=n+1n−1(nn−1−1)(nn−1+1)=n+1n+12n−1n−1=n+1⇒n=2(notsuitable,it′sastraight)case3:sin(3α)n−1=sinαn=sin(2α)n+1sin(2α)sinα=2cosα=n+1nsin(3α)sinα=(2cosα)2−1=n−1n(n+1n−1)(n+1n+1)=n−1n⇒n=−2<0(notsuitable)case4:sin(3α)n−1=sin(2α)n=sinαn+1sin(2α)sinα=2cosα=nn+1sin(3α)sinα=(2cosα)2−1=n−1n+1(nn+1−1)(nn+1+1)=n−1n+1⇒n=−2<0(notsuitable)case5:sin(2α)n−1=sinαn=sin(3α)n+1sin(2α)sinα=2cosα=n−1nsin(3α)sinα=(2cosα)2−1=n+1n(n−1n+1)(n−1n−1)=n+1nn=−2<0(notsuitable)case6:sin(2α)n−1=sin(3α)n=sinαn+1sin(2α)sinα=2cosα=n−1n+1sin(3α)sinα=(2cosα)2−1=nn+1(n−1n+1+1)(n−1n+1−1)=nn+1−4n+1=1⇒n=−5<0(notsuitable)⇒theonlysolutionis:4,5,6
Commented by chux last updated on 03/Aug/17
but B=3α
butB=3α
Commented by ajfour last updated on 03/Aug/17
The same here. Thank you Sir.
Thesamehere.ThankyouSir.
Commented by chux last updated on 02/Aug/17
wow..... this is amazing! i m most grateful sir.
wow…..thisisamazing!immostgratefulsir.
Commented by mrW1 last updated on 03/Aug/17
since we don′t know which angle is as double so large as which other angle, we must try out all 6 possibilities: A=α, B=2α, C=180−3α A=α, B=180−3α, C=2α A=2α, B=α, C=180−3α A=180−3α, B=α, C=2α A=2α, B=180−3α, C=α A=180−3α, B=2α, C=α note: sin (180−3α)=sin (3α)
sincewedon′tknowwhichangleisasdoublesolargeaswhichotherangle,wemusttryoutall6possibilities:A=α,B=2α,C=180−3αA=α,B=180−3α,C=2αA=2α,B=α,C=180−3αA=180−3α,B=α,C=2αA=2α,B=180−3α,C=αA=180−3α,B=2α,C=αnote:sin(180−3α)=sin(3α)
Commented by chux last updated on 03/Aug/17
i′ve understood it now... its clear sir.
i′veunderstooditnow…itsclearsir.
Commented by chux last updated on 04/Aug/17
thanks sir
thankssir

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