Angle Subtended by a Chord at a Point: We may have come across many objects in daily life, which are round in shape, such as wheels of a vehicle, bangles, dials of clocks, coins, keyrings, buttons of shirts, etc. A circle is a closed object, which is in a round shape.

The distance of the point on the circle to its centre is called the radius, whereas the line segment joining any two points on the boundary or circumference of a circle is called the chord of a circle. Various angles can be formed in a circle joining the endpoints of the chords, and those angles are called subtended angles.

Chord of a Circle

The collection of all the points in a plane, which are at a fixed distance from a fixed point in the plane, is called a circle. The various terms associated with the circle are discussed below:

Radius: The fixed distance between the centre and any point on the circle.

Diameter: The longest chord of the circle that passes through the centre.

Chord: The line segment joining any two points on the circle.

Tangent: The line touching the circle at only one point.

Arc: The part of the boundary of a circle.

The line segment joining any two points on the boundary or the circumference of a circle is called the chord of a circle. A circle can have infinite chords. The diameter is the longest chord of the circle.

In the above figure, the line segment $QR$ is the chord of a circle, as it is joining the two points $Q$ and $R$ on the circle.

Angle Subtended by a Chord at a Point

Take a line segment $PQ$ and a point $R$ not on the line containing $PQ$. Join $PR$ and $QR$. Then $\mathrm{\angle }PRQ$ is called the angle subtended by the line segment $PQ$ at the point $R$.

Look at the below-given figure. What are angles $POQ,PRQ$ and $PSQ$ called?

Here, $\mathrm{\angle }POQ$ is the angle subtended by the chord $PQ$ at the centre $O,\mathrm{\angle }PRQ$ and $\mathrm{\angle }PSQ$ are the angles subtended by $PQ$ at points $R$ and $S$ on the major and minor arcs $PRQ$ and $ASQ$ respectively.

Let us discuss the relationship between the chords and the angle subtended by the chord at various points on a circle.

Theorem:
Equal chords of a circle subtend equal angles at the centre.

Proof:
Consider a circle with centre $O$

Let us assume the chords of the circles $AB$ and $CD$ are equal.

$AB=CD\dots \dots ..(i)$

In triangles $AOB$ and $COD$,

$OA=OC$ (Radius of the same circle)

$OB=OD$ (Radius of the same circle)

$AB=CD$ (Given equal chords)

By the $SSS$ congruency rule,

$\mathrm{\Delta }AOB\cong \mathrm{\Delta }COD$

We know that corresponding parts of congruent triangles are equal.

So, $\mathrm{\angle }AOB=\mathrm{\angle }COD$.

Therefore, equal chords subtend equal angles at the centre of a circle.

Converse Theorem

The converse of the above theorem states that If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.

Proof:
Consider a circle with centre $O$

Let us assume the chords of the circles $AB$ and $CD$ subtend equal angles at the centre.

$\mathrm{\angle }AOB=\mathrm{\angle }COD\dots ..(i)$

In triangles $AOB$ and $COD$,

$OA=OC$ (Radius of the circle)

$OB=OD$ (Radius of the circle)

$\mathrm{\angle }AOB=\mathrm{\angle }COD$ (given that these two angles are equal)

By the $SAS$ congruency rule, we have $\mathrm{\Delta }AOB\cong \mathrm{\Delta }COD$

We know that corresponding parts of congruent triangles are equal.

So, $AB=CD$.

Therefore, If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.

Angle Subtended by a Chord at the Centre and Any Other Point on the Circle

Join $O$ and $A$ and extends the ray to $B$.

In $\mathrm{\Delta }ABQ$

$\mathrm{\angle }BOQ=OAQ+\mathrm{\angle }OQA\dots \dots \dots (i)$ (Exterior angle)

In $\mathrm{\Delta }AOQ$

$OA=OQ$ (Radius)

Thus, $\mathrm{\Delta }AOQ$ is an isosceles triangle.

So, $\mathrm{\angle }OAQ=\mathrm{\angle }OQA\dots \dots \dots (ii)$

From $(i)$ and $(ii)$

$\mathrm{\angle }BOQ=2\mathrm{\angle }OAQ\dots \dots (iii)$

Similarly, $\mathrm{\angle }BOP=2\mathrm{\angle }OAP\dots \dots (iv)$

Adding $(iii)$ and $(iv)$

$\mathrm{\angle }BOQ+\mathrm{\angle }BOP=2\mathrm{\angle }OAP+2\mathrm{\angle }OAQ$

From the figure,

$\mathrm{\angle }POQ=2\mathrm{\angle }PAQ$

Therefore, the angle subtends by the chord at the centre $\left(\mathrm{\angle }POQ\right)$ equals twice the angle subtended at the circumference $\left(\mathrm{\angle }PAQ\right)$.

Angles Subtended by the Chord at Points on the Circle

Students can learn about angles subtended by chord properties, and theorems related to Angles Subtended by the Chord at Points on the Circle below:

Angles in the same segments of a circle are equal. In another way, we can say that a chord subtends equal angles at any part of the circle’s circumference.

Consider the circle as shown above, in which chord $PQ$ subtends angles $PAQ$ and $PCQ$ at any two points $A$ and $C$ on the circumference of a circle.

We know that the angle subtended by the chord at the centre is double the angle subtended by it at any point on the circle.

Thus, $\mathrm{\angle }POQ=2\mathrm{\angle }PAQ\dots \dots .(i)$

Similarly, $\mathrm{\angle }POQ=2\mathrm{\angle }PCQ\dots \dots .(ii)$

From $(i)$ and $(ii)$,

$2\mathrm{\angle }PAQ=2\mathrm{\angle }PCQ$

$\mathrm{\angle }PAQ=\mathrm{\angle }PCQ$

Thus, angles made by the chord at the same segments are equal.

Alternate Segment Theorem

For a circle, the angle made between the chord a tangent and at the point of contact is equal to the angle made by the chord in the alternate segment.

Suppose at a point of contact $P$, the tangent is drawn to the circle, and the chord $PQ$ is drawn through the point $P$ which is inclined at an angle $\alpha$ to the tangent, and it subtends an angle $\beta$ at a point $R$ on the boundary of the circle as shown below:

Let the circle has the centre $O$. $AB$ is the tangent drawn on the circle at the point of contact $P$, and $PQ$ is the chord drawn between the points $P$ and $Q$

$\alpha$ be the angle made by the chord with the tangent at the point of contact, and $\beta$ be the angle made by the chord at a point $R$ on the circle.

In the triangle $OPQ$, we have,

$OP=OQ$ (Radius of the same circle)

We know that angles opposite to equal sides are equal.

Thus, $\mathrm{\angle }OPQ=\mathrm{\angle }OQP\dots .(i)$

By angle sum property, $\mathrm{\angle }POQ+\mathrm{\angle }OPQ+\mathrm{\angle }OQP={180}^{\mathrm{o}}$

$\mathrm{\angle }POQ+{180}^{\mathrm{o}}–2\mathrm{\angle }OPQ\dots \dots (ii)$

 We know that radius is perpendicular to the tangent at a point of contact.

$\Rightarrow \mathrm{\angle }OPB={90}^{\mathrm{o}}=\mathrm{\angle }OPQ+\alpha$

$\mathrm{\angle }OPQ={90}^{\mathrm{o}}–\alpha \dots ..(iii)$

From $(ii)$ and $(iii)$,

$\mathrm{\angle }POQ=2\alpha$

We know that the angle subtended by the chord at the centre of the circle is twice the angle subtended at any point on the circle’s circumference.

$\mathrm{\angle }POQ=2\mathrm{\angle }PRQ$

$\Rightarrow 2\alpha =2\beta$

$\Rightarrow \alpha =\beta$

Hence, proved.

Solved Examples on Angle Subtended by a Chord at a Point

Q.1. Find the angles $x$ and $y$ in the below-given figure, such that $BC$ is the diameter, $AC$ is the chord of a circle, and $XY$ is the tangent drawn at the point of contact $C$.

Ans: $AC$ is the chord of a circle, and $XY$ is the tangent drawn at the point of contact $C$

Given, $\mathrm{\angle }CAB={90}^{\mathrm{o}}$ and $\mathrm{\angle }ACB={30}^{\mathrm{o}}$.

In a triangle $ABC$ by angle sum property of a triangle,

$\Rightarrow \mathrm{\angle }ABC+\mathrm{\angle }BAC+\mathrm{\angle }ACB={180}^{\mathrm{o}}$

$\Rightarrow y+{90}^{\mathrm{o}}+{30}^{\mathrm{o}}={180}^{\mathrm{o}}$

$\Rightarrow y+{120}^{\mathrm{o}}={180}^{\mathrm{o}}$

$\Rightarrow y={180}^{\mathrm{o}}–{120}^{\mathrm{o}}$

$\Rightarrow y={60}^{\mathrm{o}}$

Here, chord $AC$ makes an angle $x$ with the tangent at a point of contact $C$ and makes an angle $y$ at point $B$ on the circle.

We know that the angle made between the chord and a tangent at the point of contact is equal to the angle made by the chord in the alternate segment.

$x=y={60}^{\mathrm{o}}$

Q.2. Prove that equal chords will make equal angles at the centre.

Ans: Let $AB$ and $CD$ are equal chords as shown below:

In triangles, $AOB$ and $COD$,

$OA=OC$ (Radius of the same circle)

$OB=OD$ (Radius of the same circle)

$AB=CD$ (Given equal chords)

By the $SSS$ congruency rule, we have,

$\mathrm{\Delta }AOB\cong \mathrm{\Delta }COD$

We know that corresponding parts of congruent triangles are equal.

 $\mathrm{\angle }AOB=\mathrm{\angle }COD$.

Therefore, equal chords will make equal angles at the centre.

Q.3. In the circle given below, $\mathrm{\angle }AOB=\mathrm{\angle }COD$ prove that $AB=CD$.

Ans: In triangles, $AOB$ and $COD$,

$OA=OC$ (Radius of the same circle)

$OB=OD$ (Radius of the same circle)

$\mathrm{\angle }AOB=\mathrm{\angle }COD$ (Given that these two angles are equal)

By the $SAS$ congruency rule,

$\mathrm{\Delta }AOB\cong \mathrm{\Delta }COD$

We know that corresponding parts of congruent triangles are equal.

$AB=CD$

Hence, proved.

Q.4. Find the value of $x$, in terms of $y$ as shown in the figure.

Ans: In the figure, chord $AB$ subtends an angle $\mathrm{\angle }ACB=x$ at the centre, and it subtends an angle $\mathrm{\angle }ADB=y$ at the circumference of the circle.

We know that angle made by the chord at the centre is twice the angle made by it at any point on the circle.

Thus, $\mathrm{\angle }ACB=2\mathrm{\angle }ADB$

$\Rightarrow x=2y$

Q.5. Find the value of the chord $RS$?

Ans: Here, chords $PQ$ and $RS$ make an equal angle $\left({30}^{\mathrm{o}}\right)$ at the centre.

We know that If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.

Thus, the length of $PQ$ and $RS$ are equal.

Given $PQ=5$ units.

Therefore, the length of $RS=5$ units.

Summary

In this article, we have studied the definitions of a circle and the circle’s radius, diameter, chord, arc, and tangent. We have discussed the definition of chord and the angles subtended by a chord at a point. We have discussed the theorems such as equal chords making equal angles at the centre and its converse etc. This article gives the solved examples, which help us solve the concepts mathematically and easily understand them.

FAQs on Angle Subtended by a Chord at a Point

Q.1. Are angles subtended by equal chords are equal?
Ans: Yes, equal chords subtend equal angles at the centre of a circle.

Q.2. What is the angle subtended by the chord at a point?
Ans: The angle made by the line segments from the chord’s endpoints to the centre or at any point on the circle, then the angle formed is called the angle subtended by the chord at a point.

Q.3. What is the angle between the tangent and the chord?
Ans: The angle formed between the tangent and the chord is equal to the angle subtended by the chord at any point on the circle (any segment).

Q.4. What is the chord-tangent (alternate segment) theorem?
Ans: For a circle, the angle made between the chord a tangent at the point of contact is equal to the angle made by the chord in the alternate segment.

Q.5. Do equal arcs subtend equal chords?
Ans: If any two chords of a circle are equal, then their corresponding arcs of the circle are the same, and conversely, we can say that if two arcs are the same, then their corresponding chords are equal.