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.

Answer 1

Hint: Use the rule of 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.

Complete step-by-step answer:
$OA = OB = AB$(Given)
Since, all the sides of the triangle are equal
$\Delta OAB$is an equilateral triangle, and we know that in an equilateral triangle all the angles are equal to ${60^ \circ }$.
Therefore, $\angle AOB = {60^ \circ }$
There is a rule which states 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.
Therefore,
$\dfrac{1}{2} \times {60^ \circ } = {30^ \circ }$
Now, since $ADBC$is a cyclic quadrilateral.
Therefore the sum of the opposite angles will be equal to${180^ \circ }$.
Therefore,
$\angle ADB + \angle ACB = {180^ \circ }$
Again applying the rule which states that the sum of either pair of opposite angles of a cyclic quadrilateral is${180^ \circ }$.
Therefore,
$\angle ADB + {30^ \circ } = {180^ \circ }$
Sending the angles on one side, we get,
$\angle ADB = {180^ \circ } - {30^ \circ }$
On further solving,
Answer =$\angle ADB = {150^ \circ }$
Note: Make sure to take the correct values in the equation while using the above rule.