Question

The chord of a circle is equal to its radius. The angle subtended by this chord at the minor arc of the circle, is :

• A. \(60^\circ\)
• B. \(75^\circ\)
• C. \(120^\circ\)
• D. \(150^\circ\)

Hint:

1. Chord = Radius \(\implies\) Triangle formed by chord and two radii to its ends is equilateral. Angle at center (\(\angle AOB\)) = \(60^\circ\). 2. Angle at major arc (\(\angle ACB\)) = \( \frac{1}{2} \times \text{angle at center} = \frac{1}{2} \times 60^\circ = 30^\circ \). 3. Points on major arc and minor arc with chord ends form a cyclic quadrilateral. Angle at minor arc (\(\angle ADB\)) + Angle at major arc (\(\angle ACB\)) = \(180^\circ\). So, Angle at minor arc = \(180^\circ - 30^\circ = 150^\circ\).

Answer 1

The Correct Option is D

Concept: This problem involves the relationship between the angle subtended by a chord at the center of a circle and the angles subtended by the same chord at points on the major and minor arcs. Step 1: Angle subtended by the chord at the center Let the circle have center O and radius \(r\). Let AB be a chord such that its length is equal to the radius, i.e., \(AB = r\). Consider \(\triangle OAB\). We have \(OA = r\) (radius), \(OB = r\) (radius), and \(AB = r\) (given). Since all three sides are equal (\(OA = OB = AB = r\)), \(\triangle OAB\) is an equilateral triangle. The angle subtended by the chord AB at the center O is \(\angle AOB\). In an equilateral triangle, all angles are \(60^\circ\). So, \(\angle AOB = 60^\circ\). Step 2: Angle subtended at the major arc The angle subtended by an arc (or chord) at the center is double the angle subtended by it at any point on the remaining part of the circle (the major arc in this case). Let C be any point on the major arc. Then, \(\angle ACB = \frac{1}{2} \angle AOB\). \[ \angle ACB = \frac{1}{2} \times 60^\circ = 30^\circ \] Step 3: Angle subtended at the minor arc Let D be any point on the minor arc. The points A, C, B, D in order form a cyclic quadrilateral ACBD. In a cyclic quadrilateral, the sum of opposite angles is \(180^\circ\). So, \(\angle ACB + \angle ADB = 180^\circ\). We found \(\angle ACB = 30^\circ\). Therefore, \(\angle ADB = 180^\circ - \angle ACB = 180^\circ - 30^\circ = 150^\circ\). The angle \(\angle ADB\) is the angle subtended by the chord AB at a point on the minor arc. Thus, the angle subtended by this chord at the minor arc of the circle is \(150^\circ\).