Lesson 2, Topic 4
In Progress

# Angle Subtended by a Chord at the Centre

Lesson Progress
0% Complete

Theorem: The angle subtended by the chord at the centre is double the angle subtended by it at any point on the circle.

Given: Circle PAQ with centre O, a chord PQ.

To prove:𝑃𝑂𝑄 = 2∠𝑃𝐴𝑄

Construction: Join 𝑂 and 𝐴 and extend the ray to 𝐵 where it touches the chord PQ.

### Proof:

In Δ𝐴𝐵𝑄

𝐵𝑂𝑄 = ∠𝑂𝐴𝑄 + ∠𝑂𝑄𝐴………(𝑖) (Exterior angle)

In Δ𝐴𝑂𝑄

𝑂𝐴 = 𝑂𝑄 (radii of the same circle)

Thus, Δ𝐴𝑂𝑄 is an isosceles triangle.

So, ∠𝑂𝐴𝑄 = ∠𝑂𝑄𝐴………(𝑖𝑖)

Equate (𝑖) and (𝑖𝑖)

𝐵𝑂𝑄 = ∠𝑂𝐴𝑄 + ∠𝑂𝐴𝑄

From (𝑖) and (𝑖𝑖)

𝐵𝑂𝑄 = 2∠𝑂𝐴𝑄……(𝑖𝑖𝑖)

Similarly, ∠𝐵𝑂𝑃 = 2∠𝑂𝐴𝑃……(𝑖𝑣)

𝐵𝑂𝑄 + ∠𝐵𝑂𝑃 = 2∠𝑂𝐴𝑃 + 2∠𝑂𝐴𝑄

∴ ∠𝐵𝑂𝑄 + ∠𝐵𝑂𝑃 = 2(𝑂𝐴𝑃 + ∠𝑂𝐴𝑄)

But ∠𝐵𝑂𝑄 + ∠𝐵𝑂𝑃 = ∠𝑃𝑂𝑄

and ∠𝑂𝐴𝑃 + ∠𝑂𝐴𝑄 = ∠𝑃𝐴𝑄

From the figure,

𝑃𝑂𝑄 = 2∠𝑃𝐴𝑄

Therefore, the angle subtends by the chord at the centre (∠𝑃𝑂𝑄) equals twice the angle subtended at the circumference (∠𝑃𝐴𝑄).

### Example 1

Find the value of 𝑥, in terms of 𝑦 as shown in the figure below.

Solution:

𝐴𝐶𝐵 = 2∠𝐴𝐷𝐵 (angle at the centre is twice the angle at the circumference)

x = 2y

### Example 2

In the diagram below, O is the centre of the circle through points L, M and N, if ∠MLN = 74º and ∠MNL = 39º, calculate ∠LON.

Solution:

∠LMN = 180° – (74° + 39°) (sum of angles in a triangle)

∠LMN = 67°

Let’s redraw the diagram by joining radii $$\scriptsize \overline{OL}$$ and $$\scriptsize \overline{ON}$$ to centre O.

∠LON = 2 x 67° (angle subtended at the centre is twice the angle at the circumference)

∠LON = 134°

### Example 3

In the diagram below, O is the centre of the circle PQRS. PQR = 84°.

Find
(i) reflex ∠POR
(ii) ∠PSR

Solution:

(i) reflex ∠POR

Remember: A reflex angle is greater than 180° while an obtuse angle is greater than 90° but less than 180°

Obtuse ∠POR = 2 x 84° (angle subtended at the centre is 2 x the angle at the circumference)

Obtuse ∠POR = 168°

Reflex ∠POR = 360° – 168° (sum of angles at a point = 360°)

Reflex ∠POR = 192°

(ii) ∠PSR

∠PSR = $$\frac{1}{2} \scriptsize \: \times \: reflex \: \angle POR$$

(angle subtended at the centre is 2 x the angle at the circumference)

∠PSR = $$\frac{1}{2} \scriptsize \: \times \: 192^o$$

∠PSR = 96°

### Example 3

Below is part of a spider web with a hole in it. Points P, Q, and R all lie on the circumference of a circle, with centre C.
∠PRC = 25°
Calculate the angle, a, of ΔPQR.

Solution:

Consider ΔPCR

$$\scriptsize \overline{CP} = \overline{CR}$$ (radii of the same circle)

Thus, ΔPCR is an isosceles triangle.

∠RPC = ∠PRC = 25° (base angles of isosceles ΔPCR)

∠PCR + ∠RPC + ∠PCR = 180° (sum of angles in ΔPCR)

∠PCR + 25° + 25° = 180°

∠PCR = 180° – 50°

∠PCR = 130°

But ∠PCR = 2 x ∠PQR (angle subtended at the centre is 2 x the angle at the circumference)

∠PQR = $$\frac{1}{2} \scriptsize \: \times \: \angle PCR$$

∠PQR = $$\frac{1}{2} \scriptsize \: \times \: 130$$

∠PQR = 65°

a = ∠PQR = 65°

#### Responses

error: Alert: Content selection is disabled!!