On a circle, ten points $A_{1}, A_{2}, A_{3}, \ldots, A_{10}$ are equally spaced. If $C$ is the centre of the circle, what is the size, in degrees, of the angle $A_{1} A_{5} C$ ?
(A) 18
(B) 36
(C) 10
(D) 72
(E) 144

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-152.jpg?height=480&width=480&top_left_y=1088&top_left_x=1430)
Solution

Join $A_{1} A_{5}, A_{1} C$, and $A_{5} C$, as shown.

Since the points $A_{1}, A_{2}, A_{3}, \ldots, A_{10}$ are equally spaced, they generate equal angles at $C$, each of measure $\frac{360}{10}=36$.

Thus, $\angle A_{1} C A_{5}=4(36)$

$$
=144
$$

Since $A_{1} C=A_{5} C$ (radii), then $\triangle A_{1} C A_{5}$ is isosceles and

$$
\angle A_{1} A_{5} C=\frac{(180-144)}{2}
$$

$$
=18
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-314.jpg?height=477&width=463&top_left_y=1347&top_left_x=1362)

Thus, the value of $\angle A_{1} A_{5} C$ is 18 degrees.

ANSWER: (A)

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