Start with [math]\Delta[/math]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][math]sinJ=\frac{a}{CJ}=\frac{a}{2R}[/math] [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 [math]sin\theta=sin\left(180-\theta\right)[/math] --> [math]sinJ=sinA[/math] in both figures. Therefore, [math]sinA=\frac{a}{2R}[/math]-->[math]\frac{a}{sinA}=2R[/math][br]The same procedure applied to the other angles of [math]\Delta[/math]ABC yields[br][math]\frac{b}{sinB}=2R[/math] and [math]\frac{c}{sinC}=2R[/math][br]Combining these results we get the extended Law of Sines,[br][i]For a triangle ABC with circumradius R,[br][math]\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=2R[/math][/i]