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.

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Solution

## The chord $AB$is equal to the radius of the circle. $OA$and $OB$are the two radii of the circle.From $\Delta OAB$. $AB=OA=OB=$ radius of the circle. $\Delta OAB$is an equilateral triangle.$\therefore AOC=60°$And,$\angle ACB=\frac{1}{2}\angle AOB$So, $\angle ACB=\frac{1}{2}×60°=30°$Now, $ACBD$is a cyclic quadrilateral,$\angle ADB+\angle ACB=180°$(Since they are the opposite angles of a cyclic quadrilateral)So, $\angle ADB=180°-30°=150°$Hence, the angle subtended by the chord at a point on the minor arc and also at a point on the major arc is $150°&30°$ respectively.

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