Triangle Inside Circle

Question

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 ?

Answer 1

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

Answer 2

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

Answer 3

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 –