Question

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

• A. ${150}^{\circ },{30}^{\circ }$
• B. ${120}^{\circ },{60}^{\circ }$
• C. ${30}^{\circ },{150}^{\circ }$
• D. ${60}^{\circ },{120}^{\circ }$

Answer 1

The correct option is C ${30}^{\circ },{150}^{\circ }$


In $\Delta OAB$,
AB = OA = OB = radius
$\therefore \Delta OAB$ is an equilateral triangle.
Therefore, each interior angle of this triangle will be of ${60}^{\circ }$.
$\Rightarrow \angle AOB={60}^{\circ }$.
$\angle ACB=\frac{1}{2}\angle AOB$
$=\frac{1}{2}\left({60}^{\circ }\right)={30}^{\circ }$
In cyclic quadrilateral ACBD,
$\angle ACB+\angle ADB={180}^{\circ }$
(Opposite angle in cyclic quadrilateral are supplementary)
$\therefore \angle ADB={180}^{\circ }-{30}^{\circ }={150}^{\circ }$
Therefore, angle subtended by this chord at a point on the major arc and the minor arc are ${30}^{\circ }$ and ${150}^{\circ }$ respectively.