Ten points on a circle

We know that $\angle A_1CA_2 = \frac{360}{10}=36$ degrees, since the points are equally spaced and we are calculating the central angle, thus $\angle A_1CA_5= 4(36)= 144$ degrees.

We know that $\triangle A_1CA_5$ is an isosceles triangle (equal radii : $A_1C = A_5C$ )

Thus $\angle A_1A_5C = 180 -144 = \frac{36}{2} = 18$ degrees.

Join $A_7A_6C$ . Now since the circle is divided into 10 equal parts, $\angle A_7C_6 = \dfrac{360}{10} = 36$ .

Join $A_1CA_5$ , now the angle $\angle A_1CA_5$ is divided into 4 equal anngles of $\angle 36$ each. Therefore,

$\angle A_1CA_5 = 4 \times 36 = 144$

Hence, $\Delta A_1C_A5$ is isoceles, $CA_1 = CA_5$ .

$\therefore \angle A_1A_5C = 90 - \dfrac{\angle A_1CA_5}{2} \text{ (using angle sum property)} \\ \boxed{\therefore \angle A_1A_5C = 90 - 72 = 18}$

$The \space area \space of \space sector \space A_1CA_{10} \space is \space \frac{1}{10}th \space the \space area \space of \space circle \space as \space the \space 10 \space points \space are \space equally \space placed$ $Using \space Area \space of \space Sector , \space \frac{\theta}{360} \times {\pi}r^{2} \space = \space \frac{1}{10} \times {\pi}r^{2}$ $Here, \space \theta \space = \space Central \space Angle \space suspended \space by \space arc \space A_1A_{10}$ $\Rightarrow \space \theta \space =\space 36^{\circ}$ $We \space know \space , \space angle \space subtended \space by \space a \space chord \space at \space the \space centre \space is \space double \space the \space angle \space subtended \space by \space it \space in \space on \space the \space circumference \space of \space the \space circle$ $\Rightarrow \angle{A_1CA_{10}} \space = \space 2\angle{A_1A_5A_{10}}$ $\angle{A_1A_5A_{10}} \space=\space \angle{A_1A_5C}$ $36^{\circ} \space=\space 2\angle{A_1A_5C}$ $\therefore \boxed{\angle{A_1A_5C}\space=\space 18^{\circ}}$