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[parent] sines law proof (Proof)

Let $ABC$ a triangle. Let $T$ a point in the circumcircle such that $BT$ is a diameter.

\includegraphics{sineslawproof}

So $\angle A=\angle CAB$ is equal to $\angle CTB$ (they subtend the same arc). Since $\triangle CBT$ is a right triangle, from the definition of sine we get

\begin{displaymath}\sin \angle CTB =\frac{BC}{BT}=\frac{a}{2R}.\end{displaymath}

On the other hand $\angle CAB=\angle CTB$ implies their sines are the same and so

\begin{displaymath}\sin \angle CAB=\frac{a}{2R}\end{displaymath}

and therefore
\begin{displaymath}\frac{a}{\sin A}=2R.\end{displaymath}

Drawing diameters passing by $C$ and $A$ will let us prove in a similar way the relations

\begin{displaymath}\frac{b}{\sin B}=2R\quad\mbox{and}\quad\frac{c}{\sin C}=2R\end{displaymath}

and we conclude that
\begin{displaymath}\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R.\end{displaymath}

Q.E.D.



"sines law proof" is owned by drini. [ owner history (1) ]
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See Also: cosines law, triangle, sines law

Keywords:  Sine, Trigonometry, Triangle, Geometry, Angles

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Cross-references: relations, similar, implies, sine, right triangle, arc, diameter, circumcircle, point, triangle

This is version 5 of sines law proof, born on 2001-11-11, modified 2002-07-28.
Object id is 757, canonical name is SinesLawProof.
Accessed 6686 times total.

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AMS MSC: 51-00 (Geometry :: General reference works )
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fiziko: by drini on 2001-11-11 21:55:35
if you have a different proof, it would be really good to have two proofs. There are several entries that already hace more than 1 proof.
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