# sines law proof

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

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

 $\sin\angle CTB=\frac{BC}{BT}=\frac{a}{2R}.$

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

 $\sin\angle CAB=\frac{a}{2R}$

and therefore

 $\frac{a}{\sin A}=2R.$

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

 $\frac{b}{\sin B}=2R\quad\mbox{and}\quad\frac{c}{\sin C}=2R$

and we conclude that

 $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R.$

Q.E.D.

Title sines law proof SinesLawProof 2013-03-22 11:57:36 2013-03-22 11:57:36 drini (3) drini (3) 9 drini (3) Proof msc 51-00 CosinesLaw Triangle SinesLaw