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 are and also at a point on the major are.

Answer 1

A circle with centre O, a chord AB = radius of the circle C and D are points on the minor and major arcs of the circle

$\therefore \angle ACBand\angle ADB$ are formed

Now in $\Delta AOB$,

OA = OB = AB ($\because$ AB = radii of the circle)

$\therefore \Delta AOB$ is an equilateral triangle,

$\angle AOB={60}^{\circ }$

Now are AB subtends $\angle AOB$ at the centre and $\angle ADB$ at the remainder part of the circle.

$\angle ADB=\frac{1}{2}\angle AOB=\frac{1}{2}\times {60}^{\circ }={30}^{\circ }$

Now ACBD is a cyclic quadrilateral,

$\therefore \angle ADB+\angle AOB=\frac{1}{2}\times {60}^{\circ }={30}^{\circ }$

Now ACBD is a cyclic quadrilateral,

$\therefore \angle ADB=\frac{1}{2}\angle AOB=\frac{1}{2}\times {60}^{\circ }={30}^{\circ }$

Now ACBD is a cyclic quadrilateral,

$\therefore \angle ADB+\angle ACB={180}^{\circ }$

(Sum of opposite angles of the cyclic quad.)

$\Rightarrow {30}^{\circ }+\angle ACB={180}^{\circ }$

$\Rightarrow \angle ACB={180}^{\circ }$

$\Rightarrow \angle ACB={180}^{\circ }-{30}^{\circ }={150}^{\circ }$

$\therefore \angle ACB={150}^{\circ }$

Hence angles are ${150}^{\circ }$, and ${30}^{\circ }$