On a circle, fifteen points $A_{1}, A_{2}, A_{3}, \ldots, A_{15}$ are equally spaced. What is the size of angle $A_{1} A_{3} A_{7}$ ?
(A) $96^{\circ}$
(B) $100^{\circ}$
(C) $104^{\circ}$
(D) $108^{\circ}$
(E) $120^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-154.jpg?height=485&width=509&top_left_y=1690&top_left_x=1407)

Part C: Each question is worth 8 credits.
Solution

Join $A_{1}, A_{3}$ and $A_{7}$ to $O$, the centre of the circle, as shown.

Since the points $A_{1}, A_{2}, A_{3}, \ldots, A_{15}$ are evenly spaced, they generate equal angles at $O$, each of measure $\frac{360 \mathrm{r}}{15}=24 \mathrm{r}$.

Thus, $-A_{1} O A_{3}=48 \mathrm{r}$ and $-A_{3} O A_{7}=96 \mathrm{r}$.

Since $O A_{1}=O A_{3}$ (radii), $\mathrm{DA}_{1} O A_{3}$ is isosceles, and

- $A_{1} A_{3} O=\frac{(180 r-48 r)}{2}$

$=66 \mathrm{r}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-291.jpg?height=482&width=501&top_left_y=1198&top_left_x=1430)

Similarly, $\mathrm{DA}_{3} \mathrm{OA}_{7}$ is isosceles, and

- $O A_{3} A_{7}=\frac{(180 r-96 r)}{2}$

$$
=42 r
$$

- $A_{1} A_{3} A_{7}=-A_{1} A_{3} O+-O A_{3} A_{7}$

$=66 \mathrm{r}+42 \mathrm{r}$

Thus, $\quad=108 \mathrm{r}$.

ANSWER: (D)

PART C: