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Theory:

Angle at the Centre and the Circumference

Illustration:

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.

In all the three cases, $\stackrel{⌢}{\mathit{QR}}$ subtend $$\angle QOR$$ at the centre and $$\angle QPR$$ at the point $$P$$ on the circumference of the circle.

Now let us 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:

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

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

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

Based on this result, the following theorem is obtained.
Theorem: The angle subtended by an arc of the circle 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. (i.e.) $$\angle QOR = 2\angle QPR$$.
Example:
Find the unknown angle $$x$$ in the given figure if the angle subtended by the arc $\stackrel{⌢}{\mathit{QR}}$ at the centre $$O$$ is $$160^{\circ}$$.

Solution:

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

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

$$x = \frac{160^{\circ}}{2}$$

$$=$$ $$80^{\circ}$$

Therefore, the unknown angle $$x$$ is $$80^{\circ}$$.