Question

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

• A. ${150}^o$
• B. ${130}^o$
• C. ${100}^o$
• D. ${120}^o$

Answer 1

Answer: (A)

${150}^o$

In $△OAB$

$\because OA=OB=AB$ [$AB$ is the chord equal to the radius of the circle]

$\therefore △OAB$ is a equilateral triangle

$\angle AOB={60}^o$

We know that the angle subtended by an arc at the centre is double the angle subtended by it at a any point on the remaining part of the circle

$\therefore \angle ACB=\frac{1}{2}\angle AOB$

$=\frac{1}{2}\times 60$

$={30}^o$

$\because ADBC$ is a cyclic quadrilateral

$\therefore \angle ADB+\angle ACB={180}^o$ [The sum of either pair of opposite angles of a cyclic quadrilateral is ${180}^o$]

$\angle ADB+{30}^o={180}^o$

$\Rightarrow \angle ADB={180}^o-{30}^o$

$\angle ADB={150}^o$