If you draw a segment from the center of the circle to the midpoint of the chord, what special relationship does this segment have with the chord? How does that help you form a right triangle?

In the right triangle you formed, what is the angle at the center, and which side represents half of the chord? Use $\sin 60^\circ$ or the $30$-$60$-$90$ triangle ratios to find that side, then double it to get the chord length.

In a Desmos expression line, type 10*sin(60°) to represent $MB = 10\sin 60^\circ$, which is half the length of the chord.

In a new expression line, type 2*(10*sin(60°)). The numerical value Desmos displays for this expression is the length of chord $\overline{AB}$.

Let $O$ be the center of the circle, and $A$ and $B$ the endpoints of the chord. Draw radii $\overline{OA}$ and $\overline{OB}$.

• $OA = OB = 10$ (both are radii).
• The central angle $\angle AOB$ is $120^\circ$. \nSo $\triangle AOB$ is an isosceles triangle with two sides of length $10$ and included angle $120^\circ$, and the chord $\overline{AB}$ is the side opposite the $120^\circ$ angle.

Let $M$ be the midpoint of chord $\overline{AB}$. Draw $\overline{OM}$. \nKey facts:

• In a circle, the segment from the center to the midpoint of a chord is perpendicular to the chord, so $\angle OMB = 90^\circ$.
• $M$ is the midpoint, so $AM = MB = \dfrac{AB}{2}$.
• $\angle AOB = 120^\circ$, so $\angle AOM = 60^\circ$ (since $OM$ bisects the central angle). \nNow consider right triangle $\triangle OMB$:
• Hypotenuse: $OB = 10$
• Angle at $O$: $\angle MOB = 60^\circ$
• Opposite side to $60^\circ$: $MB$ (half of the chord) \nUse sine in this right triangle:

From the previous step: