# Law of Sines

Start with $\Delta$ABC and circumscribe it about a circle with center at [i]O[/i] and with radius equal to R units.[br]Draw in the diameter [i]CJ[/i] and the chord[i] BJ[/i]. <CBJ is a right angle, since it is inscribed in a semicircle.[br]Therefore in both figures,[br]$sinJ=\frac{a}{CJ}=\frac{a}{2R}$ [br]In the first diagram <J=<A because they are both inscribed in the same arc of the circle.[br]In the second diagram <J=180-<A, because opposite angles of an inscribed quadrilateral are supplementary.[br]Remember that $sin\theta=sin\left(180-\theta\right)$ --> $sinJ=sinA$ in both figures. Therefore, $sinA=\frac{a}{2R}$-->$\frac{a}{sinA}=2R$[br]The same procedure applied to the other angles of $\Delta$ABC yields[br]$\frac{b}{sinB}=2R$ and $\frac{c}{sinC}=2R$[br]Combining these results we get the extended Law of Sines,[br][i]For a triangle ABC with circumradius R,[br]$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=2R$[/i]