# 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 $x$?

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Scalene triangle is the more customary term. –  Ｊ. Ｍ. May 2 '12 at 15:26
Here's an idea: you can use the Pythagorean theorem along with the lengths you've been given to reckon the radius. The two radii and the segment of length $a$ form a triangle; use the law of cosines to find the angle opposite $a$; half of that angle is the inscribed angle you need. –  Ｊ. Ｍ. May 2 '12 at 15:29
Or use the extended law of sines: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R.$$ –  Thomas Belulovich May 2 '12 at 15:35

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, $$\arctan(\sqrt{3})=\frac{\pi}{3}.$$

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How do you know that the two right triangles are congruent? –  Cobold May 2 '12 at 15:56
@Cobold: the segment $r$ is perpendicular to the chord $a$, so it is the perpendicular bisector of the chord $a$, which is what causes the two triangles to have the same leg lengths; the Pythagorean theorem will then give the same hypotenuse lengths, so all three sides of each triangle are congruent, hence the triangles are congruent. –  Isaac May 2 '12 at 16:00
@Cobold: I should have said, the segment $r$ is perpendicular to the chord and includes the center of the circle, so it is the perpendicular bisector of the chord. –  Isaac May 2 '12 at 16:22

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 $X = 60°$

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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 $X$.