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Question Number 211609 by MATHEMATICSAM last updated on 14/Sep/24

Prove that, in a triangle the ratios of the sides and the sine of the opposite angles are equal. Also prove that each ratio is equal to the diameter of the circum circle of the triangle.

$$\mathrm{Prove}\:\mathrm{that},\:\mathrm{in}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{the}\:\mathrm{ratios}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{sides}\:\mathrm{and}\:\mathrm{the}\:\mathrm{sine}\:\mathrm{of}\:\mathrm{the}\:\mathrm{opposite}\:\mathrm{angles} \\ $$$$\mathrm{are}\:\mathrm{equal}.\:\mathrm{Also}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{each}\:\mathrm{ratio}\:\mathrm{is} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{diameter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circum}\:\mathrm{circle} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}. \\ $$

Answered by Frix last updated on 14/Sep/24

We know a^2 +b^2 −2abcos γ =c^2 ⇔ cos γ =((a^2 +b^2 −c^2 )/(2ab)) ⇒ sin γ =(√(1−cos^2 γ))= =((√((a+b+c)(a+b−c)(a+c−b)(b+c−a)))/(2ab))= =(δ/(2ab)) ⇒ sin α =(δ/(2bc))∧sin β =(δ/(2ac)) ⇒ (a/(sin α))=(b/(sin β))=(c/(sin γ))=((2abc)/δ)=((abc)/(2A)) [The area of the triangle is A=(δ/4)] The radius of the circumcircle is R=(a/(2sin α))=(b/(2sin β))=(c/(2sin γ))=((abc)/δ)=((abc)/(4A))

$$\mathrm{We}\:\mathrm{know} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\mathrm{cos}\:\gamma\:={c}^{\mathrm{2}} \\ $$$$\Leftrightarrow \\ $$$$\mathrm{cos}\:\gamma\:=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$$$\Rightarrow \\ $$$$\mathrm{sin}\:\gamma\:=\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\gamma}= \\ $$$$=\frac{\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}+{c}−{b}\right)\left({b}+{c}−{a}\right)}}{\mathrm{2}{ab}}= \\ $$$$=\frac{\delta}{\mathrm{2}{ab}} \\ $$$$\Rightarrow \\ $$$$\mathrm{sin}\:\alpha\:=\frac{\delta}{\mathrm{2}{bc}}\wedge\mathrm{sin}\:\beta\:=\frac{\delta}{\mathrm{2}{ac}} \\ $$$$\Rightarrow \\ $$$$\frac{{a}}{\mathrm{sin}\:\alpha}=\frac{{b}}{\mathrm{sin}\:\beta}=\frac{{c}}{\mathrm{sin}\:\gamma}=\frac{\mathrm{2}{abc}}{\delta}=\frac{{abc}}{\mathrm{2}{A}} \\ $$$$\left[\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{is}\:{A}=\frac{\delta}{\mathrm{4}}\right] \\ $$$$\mathrm{The}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{is} \\ $$$${R}=\frac{{a}}{\mathrm{2sin}\:\alpha}=\frac{{b}}{\mathrm{2sin}\:\beta}=\frac{{c}}{\mathrm{2sin}\:\gamma}=\frac{{abc}}{\delta}=\frac{{abc}}{\mathrm{4}{A}} \\ $$

Answered by mr W last updated on 14/Sep/24

Commented by mr W last updated on 14/Sep/24

a=2×R sin ((∠BOC)/2)=2R sin A ⇒(a/(sin A))=2R=Diameter similarly (b/(sin B))=2R=Diameter (c/(sin C))=2R=Diameter i.e. (a/(sin A))=(b/(sin B))=(c/(sin C))=Diameter of circumcircle

$${a}=\mathrm{2}×{R}\:\mathrm{sin}\:\frac{\angle{BOC}}{\mathrm{2}}=\mathrm{2}{R}\:\mathrm{sin}\:{A} \\ $$$$\Rightarrow\frac{{a}}{\mathrm{sin}\:{A}}=\mathrm{2}{R}={Diameter} \\ $$$${similarly} \\ $$$$\frac{{b}}{\mathrm{sin}\:{B}}=\mathrm{2}{R}={Diameter} \\ $$$$\frac{{c}}{\mathrm{sin}\:{C}}=\mathrm{2}{R}={Diameter} \\ $$$${i}.{e}. \\ $$$$\frac{{a}}{\mathrm{sin}\:{A}}=\frac{{b}}{\mathrm{sin}\:{B}}=\frac{{c}}{\mathrm{sin}\:{C}}={Diameter}\:{of}\:{circumcircle} \\ $$

Commented by Ari last updated on 16/Sep/24

Mr.Why a=2Rsin((BOC)/2)=2RsinA

$${Mr}.{Why}\:{a}=\mathrm{2}{Rsin}\frac{{BOC}}{\mathrm{2}}=\mathrm{2}{RsinA} \\ $$$$ \\ $$

Commented by mr W last updated on 16/Sep/24

Commented by mr W last updated on 16/Sep/24

∠BOC=2×∠A ⇒((∠BOC)/2)=∠A DC=OC×sin ∠COD =R sin ((∠BOC)/2)=R sin ∠A BD=DC a=BC=2×DC=2R sin ∠A

$$\angle{BOC}=\mathrm{2}×\angle{A} \\ $$$$\Rightarrow\frac{\angle{BOC}}{\mathrm{2}}=\angle{A} \\ $$$${DC}={OC}×\mathrm{sin}\:\angle{COD} \\ $$$$\:\:\:\:\:\:\:={R}\:\mathrm{sin}\:\frac{\angle{BOC}}{\mathrm{2}}={R}\:\mathrm{sin}\:\angle{A} \\ $$$${BD}={DC} \\ $$$${a}={BC}=\mathrm{2}×{DC}=\mathrm{2}{R}\:\mathrm{sin}\:\angle{A} \\ $$

Commented by Ari last updated on 16/Sep/24

Thank you Mr!

$${Thank}\:{you}\:{Mr}! \\ $$

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