### Theory:

Angle at the Centre and the Circumference
Theorem: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Explanation:

The theorem states that the angle subtended by the arc $\stackrel{⌢}{\mathit{QR}}$ of the circle at the centre $$O$$ is twice the angle subtended by the point $$P$$ at any remaining part of the circle. That is, $$\angle QOR = 2\angle QPR$$.

Proof of the theorem:

Consider a circle with centre $$O$$ and place three points $$P$$, $$Q$$ and $$R$$ on the circumference of the circle such that $\stackrel{⌢}{\mathit{QR}}$ subtends $$\angle QOR$$ at the centre $$O$$ and $$\angle QPR$$ at the circumference of the circle.

Consider three cases where (i) Figure $$1$$ - $\stackrel{⌢}{\mathit{QR}}$ is minor (ii) Figure $$2$$ - $\stackrel{⌢}{\mathit{QR}}$ is a semicircle and (iii) Figure $$3$$ - $\stackrel{⌢}{\mathit{QR}}$ is a major arc.

Extend $$PO$$ to $$S$$ and join $$PS$$.

In the triangle $$POQ$$, $$\angle QOS$$ is an exterior angle.

By the property of the triangle, the exterior angle is equal to the sum  of the interior opposite angles.

That is, $$\angle QOS$$ $$=$$ $$\angle OPQ$$ $$+$$ $$\angle PQO$$.

Here, $$OP$$ $$=$$ $$OQ$$. Since the radii are equal.

This implies that the $$\angle OPQ$$ $$=$$ $$\angle PQO$$ as they form the base angles of the isosceles triangle $$POQ$$.

Therefore, $$\angle QOS$$ $$=$$ $$\angle OPQ$$ $$+$$ $$\angle OPQ$$

$$=$$ $$2 \angle OPQ$$                    $$……$$ $$\text{equation }(1)$$

In the triangle $$POR$$, $$\angle ROS$$ is an exterior angle.

By the property of the triangle, the exterior angle is equal to the sum of the interior opposite angles.

That is, $$\angle ROS$$ $$=$$ $$\angle OPR$$ $$+$$ $$\angle PRO$$.

Here, $$OP$$ $$=$$ $$OR$$. Since the radii are equal.

This implies that the $$\angle OPR$$ $$=$$ $$\angle PRO$$ as they form the base angles of the isosceles triangle $$POR$$.

Therefore, $$\angle ROS$$ $$=$$ $$\angle OPR$$ $$+$$ $$\angle OPR$$

$$=$$ $$2 \angle OPR$$                    $$……$$ $$\text{equation }(2)$$

Add equation $$(1)$$ and $$(2)$$, we get:

$$\angle QOS$$ $$+$$ $$\angle ROS$$ $$=$$ $$2 \angle OPQ$$ $$+$$ $$2 \angle OPR$$

$$\angle QOR$$ $$=$$ $$2 (\angle OPQ + \angle OPR)$$

$$\angle QOR = 2\angle QPR$$

The obtained result is true for the cases (i) and (ii) whereas for case (iii) we have reflex $$\angle QOR = 2\angle QPR$$.

Example:
Find the unknown angle $$x$$ in the given figure if the angle subtended by the major arc $\stackrel{⌢}{\mathit{QR}}$ at the centre $$O$$ is $$240^{\circ}$$.

Solution:

By the theorem, reflex $$\angle QOR = 2\angle QPR$$.

This implies, $$\angle QPR = \frac{1}{2} \times$$ reflex $$\angle QOR$$

$$x = \frac{240^{\circ}}{2}$$

$$=$$ $$120^{\circ}$$

Therefore, the unknown angle $$x$$ is $$120^{\circ}$$.