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-

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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

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

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

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

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

Commented by Ari last updated on 16/Sep/24

$$Mr.Whya=2Rsin\frac{BOC}{2}=2RsinA$$$$$$

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

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

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

Commented by Ari last updated on 16/Sep/24

$$ThankyouMr!$$

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