Question Number 211609 by MATHEMATICSAM last updated on 14/Sep/24
$$\mathrm{Prove}\:\mathrm{that},\:\mathrm{in}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{the}\:\mathrm{ratios}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{sides}\:\mathrm{and}\:\mathrm{the}\:\mathrm{sine}\:\mathrm{of}\:\mathrm{the}\:\mathrm{opposite}\:\mathrm{angles} \\ $$$$\mathrm{are}\:\mathrm{equal}.\:\mathrm{Also}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{each}\:\mathrm{ratio}\:\mathrm{is} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{diameter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circum}\:\mathrm{circle} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}. \\ $$
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))](https://www.tinkutara.com/question/Q211618.png)
$$\mathrm{We}\:\mathrm{know} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\mathrm{cos}\:\gamma\:={c}^{\mathrm{2}} \\ $$$$\Leftrightarrow \\ $$$$\mathrm{cos}\:\gamma\:=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$$$\Rightarrow \\ $$$$\mathrm{sin}\:\gamma\:=\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\gamma}= \\ $$$$=\frac{\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}+{c}−{b}\right)\left({b}+{c}−{a}\right)}}{\mathrm{2}{ab}}= \\ $$$$=\frac{\delta}{\mathrm{2}{ab}} \\ $$$$\Rightarrow \\ $$$$\mathrm{sin}\:\alpha\:=\frac{\delta}{\mathrm{2}{bc}}\wedge\mathrm{sin}\:\beta\:=\frac{\delta}{\mathrm{2}{ac}} \\ $$$$\Rightarrow \\ $$$$\frac{{a}}{\mathrm{sin}\:\alpha}=\frac{{b}}{\mathrm{sin}\:\beta}=\frac{{c}}{\mathrm{sin}\:\gamma}=\frac{\mathrm{2}{abc}}{\delta}=\frac{{abc}}{\mathrm{2}{A}} \\ $$$$\left[\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{is}\:{A}=\frac{\delta}{\mathrm{4}}\right] \\ $$$$\mathrm{The}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{is} \\ $$$${R}=\frac{{a}}{\mathrm{2sin}\:\alpha}=\frac{{b}}{\mathrm{2sin}\:\beta}=\frac{{c}}{\mathrm{2sin}\:\gamma}=\frac{{abc}}{\delta}=\frac{{abc}}{\mathrm{4}{A}} \\ $$
Answered by mr W last updated on 14/Sep/24
Commented by mr W last updated on 14/Sep/24
$${a}=\mathrm{2}×{R}\:\mathrm{sin}\:\frac{\angle{BOC}}{\mathrm{2}}=\mathrm{2}{R}\:\mathrm{sin}\:{A} \\ $$$$\Rightarrow\frac{{a}}{\mathrm{sin}\:{A}}=\mathrm{2}{R}={Diameter} \\ $$$${similarly} \\ $$$$\frac{{b}}{\mathrm{sin}\:{B}}=\mathrm{2}{R}={Diameter} \\ $$$$\frac{{c}}{\mathrm{sin}\:{C}}=\mathrm{2}{R}={Diameter} \\ $$$${i}.{e}. \\ $$$$\frac{{a}}{\mathrm{sin}\:{A}}=\frac{{b}}{\mathrm{sin}\:{B}}=\frac{{c}}{\mathrm{sin}\:{C}}={Diameter}\:{of}\:{circumcircle} \\ $$
Commented by Ari last updated on 16/Sep/24
$${Mr}.{Why}\:{a}=\mathrm{2}{Rsin}\frac{{BOC}}{\mathrm{2}}=\mathrm{2}{RsinA} \\ $$$$ \\ $$
Commented by mr W last updated on 16/Sep/24
Commented by mr W last updated on 16/Sep/24
$$\angle{BOC}=\mathrm{2}×\angle{A} \\ $$$$\Rightarrow\frac{\angle{BOC}}{\mathrm{2}}=\angle{A} \\ $$$${DC}={OC}×\mathrm{sin}\:\angle{COD} \\ $$$$\:\:\:\:\:\:\:={R}\:\mathrm{sin}\:\frac{\angle{BOC}}{\mathrm{2}}={R}\:\mathrm{sin}\:\angle{A} \\ $$$${BD}={DC} \\ $$$${a}={BC}=\mathrm{2}×{DC}=\mathrm{2}{R}\:\mathrm{sin}\:\angle{A} \\ $$
Commented by Ari last updated on 16/Sep/24
$${Thank}\:{you}\:{Mr}! \\ $$