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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.
Given:
A chord of a circle is equal to the radius of the circle.
To do:
We have to find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Solution:
In the figure, AB is equal to the radius of the circle.
In $\triangle OAB$,
$OA=OB=AB$ (Radii of the circle)
Therefore,
$\triangle OAB$ is an equilateral triangle.
$\angle AOC=60^o$
$\angle ∠ACB= \frac{1}{2}\angle AOB=\frac{1}{2}\times60^o=30^o$
ACBD is a cyclic quadrilateral.
This implies,
$\angle ACB+\angle ADB=180^o$ (Opposite angles of a cyclic quadrilateral are supplementary)
$\angle ADB=180^o−30^o=150^o$.
Therefore, angles subtended by the chord at a point on the minor arc and at a point on the major arc are $150^o$ and $30^o$ respectively.
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