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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.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{side}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle} \\ $$$$\mathrm{if}\:\mathrm{side}\:\mathrm{lengths}\:\mathrm{are}\:\mathrm{consecutive}\: \\ $$$$\mathrm{integers},\mathrm{and}\:\mathrm{one}\:\mathrm{of}\:\mathrm{whose}\:\mathrm{angles} \\ $$$$\mathrm{is}\:\mathrm{twice}\:\mathrm{as}\:\mathrm{large}\:\mathrm{as}\:\mathrm{another}. \\ $$

Commented by chux last updated on 02/Aug/17

please help

$$\mathrm{please}\:\mathrm{help} \\ $$

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

$$\mathrm{a}=\mathrm{n}−\mathrm{1} \\ $$$$\mathrm{b}=\mathrm{n} \\ $$$$\mathrm{c}=\mathrm{n}+\mathrm{1} \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{1}: \\ $$$$\mathrm{A}=\alpha \\ $$$$\mathrm{C}=\mathrm{2}\alpha \\ $$$$\mathrm{B}=\mathrm{180}−\mathrm{3}\alpha \\ $$$$\frac{\mathrm{sin}\:\mathrm{A}}{\mathrm{a}}=\frac{\mathrm{sin}\:\mathrm{B}}{\mathrm{b}}=\frac{\mathrm{sin}\:\mathrm{C}}{\mathrm{c}} \\ $$$$\frac{\mathrm{sin}\:\alpha}{\mathrm{n}−\mathrm{1}}=\frac{\mathrm{sin}\:\left(\mathrm{3}\alpha\right)}{\mathrm{n}}=\frac{\mathrm{sin}\:\left(\mathrm{2}\alpha\right)}{\mathrm{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{2}\alpha\right)}{\mathrm{sin}\:\alpha}=\mathrm{2cos}\:\alpha=\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{3}\alpha\right)}{\mathrm{sin}\:\alpha}=\frac{\mathrm{3}\:\mathrm{sin}\:\alpha−\mathrm{4}\:\mathrm{sin}^{\mathrm{3}} \:\alpha}{\mathrm{sin}\:\alpha}=\mathrm{3}−\mathrm{4sin}^{\mathrm{2}} \:\alpha=\left(\mathrm{2cos}\:\alpha\right)^{\mathrm{2}} −\mathrm{1}=\frac{\mathrm{n}}{\mathrm{n}−\mathrm{1}} \\ $$$$\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}−\mathrm{1}}\right)^{\mathrm{2}} −\mathrm{1}=\frac{\mathrm{n}}{\mathrm{n}−\mathrm{1}} \\ $$$$\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}−\mathrm{1}}+\mathrm{1}\right)\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}−\mathrm{1}}−\mathrm{1}\right)=\frac{\mathrm{n}}{\mathrm{n}−\mathrm{1}} \\ $$$$\frac{\mathrm{4}}{\mathrm{n}−\mathrm{1}}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{n}=\mathrm{5} \\ $$$$\Rightarrow\mathrm{sides}\:\mathrm{of}\:\mathrm{triangle}\:=\mathrm{4},\mathrm{5},\mathrm{6} \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{2}: \\ $$$$\frac{\mathrm{sin}\:\alpha}{\mathrm{n}−\mathrm{1}}=\frac{\mathrm{sin}\:\left(\mathrm{2}\alpha\right)}{\mathrm{n}}=\frac{\mathrm{sin}\:\left(\mathrm{3}\alpha\right)}{\mathrm{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{2}\alpha\right)}{\mathrm{sin}\:\alpha}=\mathrm{2cos}\:\alpha=\frac{\mathrm{n}}{\mathrm{n}−\mathrm{1}} \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{3}\alpha\right)}{\mathrm{sin}\:\alpha}=\left(\mathrm{2cos}\:\alpha\right)^{\mathrm{2}} −\mathrm{1}=\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}−\mathrm{1}} \\ $$$$\left(\frac{\mathrm{n}}{\mathrm{n}−\mathrm{1}}−\mathrm{1}\right)\left(\frac{\mathrm{n}}{\mathrm{n}−\mathrm{1}}+\mathrm{1}\right)=\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{2n}−\mathrm{1}}{\mathrm{n}−\mathrm{1}}=\mathrm{n}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{n}=\mathrm{2}\:\left(\mathrm{not}\:\mathrm{suitable},\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{straight}\right) \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{3}: \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{3}\alpha\right)}{\mathrm{n}−\mathrm{1}}=\frac{\mathrm{sin}\:\alpha}{\mathrm{n}}=\frac{\mathrm{sin}\:\left(\mathrm{2}\alpha\right)}{\mathrm{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{2}\alpha\right)}{\mathrm{sin}\:\alpha}=\mathrm{2cos}\:\alpha=\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}} \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{3}\alpha\right)}{\mathrm{sin}\:\alpha}=\left(\mathrm{2cos}\:\alpha\right)^{\mathrm{2}} −\mathrm{1}=\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}} \\ $$$$\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}}−\mathrm{1}\right)\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}}+\mathrm{1}\right)=\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}} \\ $$$$\Rightarrow\mathrm{n}=−\mathrm{2}\:<\mathrm{0}\:\left(\mathrm{not}\:\mathrm{suitable}\right) \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{4}: \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{3}\alpha\right)}{\mathrm{n}−\mathrm{1}}=\frac{\mathrm{sin}\:\left(\mathrm{2}\alpha\right)}{\mathrm{n}}=\frac{\mathrm{sin}\:\alpha}{\mathrm{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{2}\alpha\right)}{\mathrm{sin}\:\alpha}=\mathrm{2cos}\:\alpha=\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{3}\alpha\right)}{\mathrm{sin}\:\alpha}=\left(\mathrm{2cos}\:\alpha\right)^{\mathrm{2}} −\mathrm{1}=\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}+\mathrm{1}} \\ $$$$\left(\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}}−\mathrm{1}\right)\left(\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}}+\mathrm{1}\right)=\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{n}=−\mathrm{2}<\mathrm{0}\:\left(\mathrm{not}\:\mathrm{suitable}\right) \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{5}: \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{2}\alpha\right)}{\mathrm{n}−\mathrm{1}}=\frac{\mathrm{sin}\:\alpha}{\mathrm{n}}=\frac{\mathrm{sin}\:\left(\mathrm{3}\alpha\right)}{\mathrm{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{2}\alpha\right)}{\mathrm{sin}\:\alpha}=\mathrm{2cos}\:\alpha=\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}} \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{3}\alpha\right)}{\mathrm{sin}\:\alpha}=\left(\mathrm{2cos}\:\alpha\right)^{\mathrm{2}} −\mathrm{1}=\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}} \\ $$$$\left(\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}}+\mathrm{1}\right)\left(\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}}−\mathrm{1}\right)=\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}} \\ $$$$\mathrm{n}=−\mathrm{2}<\mathrm{0}\:\left(\mathrm{not}\:\mathrm{suitable}\right) \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{6}: \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{2}\alpha\right)}{\mathrm{n}−\mathrm{1}}=\frac{\mathrm{sin}\:\left(\mathrm{3}\alpha\right)}{\mathrm{n}}=\frac{\mathrm{sin}\:\alpha}{\mathrm{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{2}\alpha\right)}{\mathrm{sin}\:\alpha}=\mathrm{2cos}\:\alpha=\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{3}\alpha\right)}{\mathrm{sin}\:\alpha}=\left(\mathrm{2cos}\:\alpha\right)^{\mathrm{2}} −\mathrm{1}=\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}} \\ $$$$\left(\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}+\mathrm{1}}+\mathrm{1}\right)\left(\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}+\mathrm{1}}−\mathrm{1}\right)=\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}} \\ $$$$\frac{−\mathrm{4}}{\mathrm{n}+\mathrm{1}}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{n}=−\mathrm{5}<\mathrm{0}\:\left(\mathrm{not}\:\mathrm{suitable}\right) \\ $$$$ \\ $$$$\Rightarrow\mathrm{the}\:\mathrm{only}\:\mathrm{solution}\:\mathrm{is}: \\ $$$$\mathrm{4},\mathrm{5},\mathrm{6} \\ $$

Commented by chux last updated on 03/Aug/17

but B=3α

$$\mathrm{but}\:\mathrm{B}=\mathrm{3}\alpha \\ $$

Commented by ajfour last updated on 03/Aug/17

The same here. Thank you Sir.

$$\mathrm{The}\:\mathrm{same}\:\mathrm{here}.\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}. \\ $$

Commented by chux last updated on 02/Aug/17

wow..... this is amazing! i m most grateful sir.

$$\mathrm{wow}.....\:\mathrm{this}\:\mathrm{is}\:\mathrm{amazing}! \\ $$$$\mathrm{i}\:\mathrm{m}\:\mathrm{most}\:\mathrm{grateful}\:\mathrm{sir}. \\ $$

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)

$$\mathrm{since}\:\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{which}\:\mathrm{angle}\:\mathrm{is} \\ $$$$\mathrm{as}\:\mathrm{double}\:\mathrm{so}\:\mathrm{large}\:\mathrm{as}\:\mathrm{which}\:\mathrm{other}\:\mathrm{angle}, \\ $$$$\mathrm{we}\:\mathrm{must}\:\mathrm{try}\:\mathrm{out}\:\mathrm{all}\:\mathrm{6}\:\mathrm{possibilities}: \\ $$$$\mathrm{A}=\alpha,\:\mathrm{B}=\mathrm{2}\alpha,\:\mathrm{C}=\mathrm{180}−\mathrm{3}\alpha \\ $$$$\mathrm{A}=\alpha,\:\mathrm{B}=\mathrm{180}−\mathrm{3}\alpha,\:\mathrm{C}=\mathrm{2}\alpha \\ $$$$\mathrm{A}=\mathrm{2}\alpha,\:\mathrm{B}=\alpha,\:\mathrm{C}=\mathrm{180}−\mathrm{3}\alpha \\ $$$$\mathrm{A}=\mathrm{180}−\mathrm{3}\alpha,\:\mathrm{B}=\alpha,\:\mathrm{C}=\mathrm{2}\alpha \\ $$$$\mathrm{A}=\mathrm{2}\alpha,\:\mathrm{B}=\mathrm{180}−\mathrm{3}\alpha,\:\mathrm{C}=\alpha \\ $$$$\mathrm{A}=\mathrm{180}−\mathrm{3}\alpha,\:\mathrm{B}=\mathrm{2}\alpha,\:\mathrm{C}=\alpha \\ $$$$ \\ $$$$\mathrm{note}:\:\mathrm{sin}\:\left(\mathrm{180}−\mathrm{3}\alpha\right)=\mathrm{sin}\:\left(\mathrm{3}\alpha\right) \\ $$

Commented by chux last updated on 03/Aug/17

i′ve understood it now... its clear sir.

$$\mathrm{i}'\mathrm{ve}\:\mathrm{understood}\:\mathrm{it}\:\mathrm{now}...\:\mathrm{its}\:\mathrm{clear} \\ $$$$\mathrm{sir}. \\ $$

Commented by chux last updated on 04/Aug/17

thanks sir

$$\mathrm{thanks}\:\mathrm{sir} \\ $$

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