# Ex.10.5 Q2 Circles Solution - NCERT Maths Class 9

## Question

A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

## Text Solution

**What is known?**

Chord’s length is equal to the radius.

**What is unknown?**

Angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

**Reasoning:**

- 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.
- A quadrilateral \(ABCD\) is called cyclic if all the four vertices of it lie on a circle.
- The sum of either pair of opposite angles of a cyclic quadrilateral is \(\begin{align}180^{\circ}.\end {align}\)

**Steps:**

Draw a circle with any radius and centre \(O\). Let \(AO\) and \(BO\) be the **\(2\)** radii of the circle and let \(AB\) be the chord equal to the length of radius. Join them to form a triangle.

\(OA = OB = AB\)

Hence \(\begin{align} \Delta {ABO} \end {align}\) becomes an equilateral triangle.

Draw **\(2\)** points \(C\) and \(D\) on the circle such that they lie on major arc and minor arc respectively.

Since \(\begin{align} \Delta {ABO} \end {align}\) is an equilateral triangle, we get \(\begin{align} \angle {ABO}=60^{\circ} \end {align}\)

For the arc \(AB,\) \(\begin {align} \angle {AOB}=2 \angle {ACB} \end {align}\) as we know that 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.

\[\begin{align}\angle ACB &= \frac{1}{2}\angle AOB\\&= \frac{1}{2} \times 60 \\&= 30^\circ \end{align}\]

As you can notice the points \(A, B, C\) and \(D\) lie on the circle. Hence \(A B C D\) is a cyclic quadrilateral.

We know that, the sum of either pair of opposite angles of a cyclic quadrilateral is \(\begin{align}180^\circ.\end {align}\)

Therefore,

\[\begin{align}\angle ACB + \angle ADB &= 180^\circ \\30 + \angle ADB &= 180\\\angle ADB &= 150^\circ \end{align}\]

So when the chord of a circle is equal to the radius of the circle, the angle subtended by the chord at a point on the minor arc is \(\begin{align}150^\circ.\end {align}\) and also at a point on the major arc is \(\begin{align}30^\circ.\end {align}\)