Points $A, B, C, D, E$, and $F$ are evenly spaced around the circle with centre $O$, as shown. The measure of $\angle A O C$ is
(A) $90^{\circ}$
(B) $150^{\circ}$
(C) $144^{\circ}$
(D) $120^{\circ}$
(E) $108^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-008.jpg?height=347&width=406&top_left_y=1783&top_left_x=1312)
Solution 1

Since $A, B, C, D, E$, and $F$ are equally spaced around the circle, moving from one point to the next corresponds to moving $\frac{1}{6}$ of the way around the circle.

Therefore, moving from $A$ to $C$ corresponds to moving $\frac{2}{6}$ (or $\frac{1}{3}$ ) of the way around the circle. Since moving around the whole circle corresponds to moving through $360^{\circ}$, then moving $\frac{1}{3}$ of the way around the circle corresponds to moving through $\frac{1}{3} \times 360^{\circ}=120^{\circ}$.

Thus, $\angle A O C=120^{\circ}$.

## Solution 2

We join $A, B, C, D, E$, and $F$ to $O$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-015.jpg?height=349&width=404&top_left_y=316&top_left_x=920)

Since $A, B, C, D, E$, and $F$ are equally spaced around the circle, the angles made at the centre by consecutive points are equal. That is,

$$
\angle A O B=\angle B O C=\angle C O D=\angle D O E=\angle E O F=\angle F O A
$$

Since these 6 angles form a complete circle, the sum of their measures is $360^{\circ}$. Therefore,

$$
\angle A O B=\angle B O C=\angle C O D=\angle D O E=\angle E O F=\angle F O A=\frac{1}{6} \times 360^{\circ}=60^{\circ}
$$

This means that $\angle A O C=\angle A O B+\angle B O C=60^{\circ}+60^{\circ}=120^{\circ}$.

ANSWER: (D)