# Triangle Inside Circle

If the radius of the circle is equal to the length of the chord $AB$, what is the value of $x$?

How would I solve this problem ?

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Is it $30^\circ$? – Quixotic Jul 18 '12 at 2:31
Yes. I am thinking I should construct and equilateral triangle from the origin and get the measure of the arc. Then use the fact that inscribed angle is double the measure of intercepted arc. Will that work ? – Rajeshwar Jul 18 '12 at 2:32
Yes, I believe you are on the right track! :) – Quixotic Jul 18 '12 at 2:34
Thanks that did the trick – Rajeshwar Jul 18 '12 at 2:40
im guessing you want an answer without trigonometry. – Carry on Smiling Jul 18 '12 at 2:41

## Without Trigonometry:

Let $O$, be the center of the circle. In $\triangle OAB$, $AB=OA=OB=$ radius implying $\triangle OAB$ to be an equilateral triangle. Thus, $\angle OAC = 10^\circ$.

Again, in $\triangle OAC$, $OA=OC$, so it is an isosceles triangle, thus $\angle OAC= \angle OCA=10^\circ$

Now, using the central angle theorem, $\angle COB = 2\times \angle BAC= 100 ^\circ$

$\triangle OBC$ is also isosceles (as $OB=OC$), thus $\angle OBC= \angle OCB=\frac 12 (180^\circ-\angle COB)=40^\circ$.

Now, $\angle OCB = 10^\circ + \angle ACB\implies \angle ACB = x = 30^\circ$

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@Chuck Fernández:The central angle theorem only holds for all angles in the same side of the segment. – Quixotic Jul 18 '12 at 3:02

The sines theorem: using your drawing: $$\frac{AB}{\sin x}=2r\,\,,\,r=\,\text{radius of the circumcircle}\Longrightarrow \frac{\rlap{/}r}{\sin x}=2\rlap{/}r\Longrightarrow$$ $$\Longrightarrow \sin x=\frac{1}{2}\Longrightarrow x=30^\circ$$

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Good answer! I thought the same all the way down. Just one point, I would like to be clarified on. How do we say that it is going to be 30 deg and not 120 deg? I guess we would have to fall back to some geometry here. – Sawarnik Oct 12 '13 at 9:12
Based only on the drawing: the side $\;AC\;$ is clearly the biggest one since it is the closest to the circle's center (remember this theorem?), thus our angle cannot be the obtuse one... – DonAntonio Oct 12 '13 at 15:57

I got the answer , by constructing an equilateral triangle from the origin.Thus making angle of the Arc 60 degrees.Now since angle is 60 degrees so the inscribed angle must be 60/2 –

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When you write "origin", do you actually mean "the circle's center"? Where are this equilateral triangle's vertices? Are these the center and A,B? If so, then you're right. – DonAntonio Jul 18 '12 at 2:47
Yes I meant the center.Its vertices are OAB – Rajeshwar Jul 18 '12 at 2:49