Lesson 2, Topic 4
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Angle Subtended by a Chord at the Centre

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

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

To prove: βˆ π‘ƒπ‘‚π‘„ = 2βˆ π‘ƒπ΄π‘„

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

Proof:

In Δ𝐴𝐡𝑄

βˆ π΅π‘‚π‘„ = βˆ π‘‚π΄π‘„ + βˆ π‘‚π‘„π΄β€¦β€¦β€¦(𝑖) (Exterior angle)

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In Δ𝐴𝑂𝑄

𝑂𝐴 = 𝑂𝑄 (radii of the same circle)

Thus, Δ𝐴𝑂𝑄 is an isosceles triangle.

So, βˆ π‘‚π΄π‘„ = βˆ π‘‚π‘„π΄β€¦β€¦β€¦(𝑖𝑖)

Equate (𝑖) and (𝑖𝑖)

βˆ π΅π‘‚π‘„ = βˆ π‘‚π΄π‘„ + βˆ π‘‚π΄π‘„

From (𝑖) and (𝑖𝑖)

βˆ π΅π‘‚π‘„ = 2βˆ π‘‚π΄π‘„β€¦β€¦(𝑖𝑖𝑖)

Similarly, βˆ π΅π‘‚π‘ƒ = 2βˆ π‘‚π΄π‘ƒβ€¦β€¦(𝑖𝑣)

Adding (𝑖𝑖𝑖) and (𝑖𝑣)

βˆ π΅π‘‚π‘„ + βˆ π΅π‘‚π‘ƒ = 2βˆ π‘‚π΄π‘ƒ + 2βˆ π‘‚π΄π‘„

∴ βˆ π΅π‘‚π‘„ + βˆ π΅π‘‚π‘ƒ = 2(βˆ π‘‚π΄π‘ƒ + βˆ π‘‚π΄π‘„)

But βˆ π΅π‘‚π‘„ + βˆ π΅π‘‚π‘ƒ = βˆ π‘ƒπ‘‚π‘„

and βˆ π‘‚π΄π‘ƒ + βˆ π‘‚π΄π‘„ = βˆ π‘ƒπ΄π‘„

Screenshot 2022 06 07 at 02.10.41

From the figure,

βˆ π‘ƒπ‘‚π‘„ = 2βˆ π‘ƒπ΄π‘„

Therefore, the angle subtends by the chord at the centre (βˆ π‘ƒπ‘‚π‘„) equals twice the angle subtended at the circumference (βˆ π‘ƒπ΄π‘„).

Example 1

Find the value of π‘₯, in terms of 𝑦 as shown in the figure below.

Screenshot 2022 06 07 at 02.51.03

Solution:

∠𝐴𝐢𝐡 = 2∠𝐴𝐷𝐡 (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.

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Solution:

∠LMN = 180Β° – (74Β° + 39Β°) (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.

Screenshot 2022 06 07 at 03.21.14

∠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:

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(i) reflex ∠POR

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

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

Obtuse ∠POR = 168°

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

Reflex ∠POR = 192°

(ii) ∠PSR

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∠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 Ξ”PQR.

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Solution:

Consider Ξ”PCR

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

Thus, Ξ”PCR is an isosceles triangle.

∠RPC = ∠PRC = 25Β° (base angles of isosceles Ξ”PCR)

∠PCR + ∠RPC + ∠PCR = 180Β° (sum of angles in Ξ”PCR)

∠PCR + 25° + 25° = 180°

∠PCR = 180Β° – 50Β°

∠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°

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