angle of an inscribed triangle I have a scalene triangle inscribed in a circle, one of its sides $a$ is $2\sqrt3$ and the length $r$ from that side to the center is $1$. I need to find the angle $x$ opposite to the side given. Here's how it looks:

How to find $x$?
 A: Draw in the radii from the center to the endpoints of $a$ to form two congruent right triangles, sharing a leg of length $r=1$ and each having another leg of length $\sqrt{3}$.  From right triangle trigonometry, the angles in these right triangles at the center of the circle has measure $\arctan(\sqrt{3})=\frac{\pi}{3},$ so the measure of the entire central angle (angle at the center of the circle) that subtends the same arc as the chord $a$ is $2\arctan(\sqrt{3})=\frac{2\pi}{3},$ and the measure of the inscribed angle $x$ is half of that, $$\arctan(\sqrt{3})=\frac{\pi}{3}.$$
A: http://www11.0zz0.com/2012/05/02/15/202577206.jpg
by using construction in the inscribed triangle 
of the center $M$
draw $ML$ & $MN$
since $r$ perpendicular on $a$ 
since $LM = NM$ >> radius
& $r$ is a common side
therefore triangle $\triangle LMO$ congrant with triangle $\Delta NMO$
by using Pythagoras theorem 
$r= 1$
$LO = \frac{1}{2}a$
$LM^2 = ON^2+r^2$ 
at $ON^2 = 3$
therefore $LM=2$
since side opposite to $30°$ half the hypotunes 
therefore angle $\angle MNO = 30°$ and 
angle $\angle MLO = 30°$ too!
IN triangle $\triangle LMO$
angle $\angle M = 180° - ( 30°+30° ) = 120°$
since angle $X$ half the centric angle 
therefore angle $X = 60°$
A: This is a job for the the inscribed angle theorem :)
Apply it to the triangle formed by connecting the endpoints of $A$ to the center and the triangle formed by the endpoints of $A$ with the vertex at $X$.
