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Question
The chord of a circle is equal to its radius. The angle subtended by this chord at the minor arc of the circle is
Options
60°
75°
120°
150°
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Solution
150°
We are given that the chord is equal to its radius.
We have to find the angle subtended by this chord at the minor arc.
We have the corresponding figure as follows:
We are given that
AO = OB = AB
So ,
\[\bigtriangleup\] AOB is an equilateral triangle.
Therefore, we have
∠AOB = 60°
Since, the angle subtended by any chord at the centre is twice of the angle subtended at any point on the circle.
So `angleAQB =(angleAOB)/2`
`= 60/2 = 30°`
Take a point P on the minor arc.
Since `square APBQ` is a cyclic quadrilateral
So, opposite angles are supplementary. That is
`angle APB + angleAQB = 180°`
`angle APB + 30° = 180°`
`angleAPB = 180° - 30°`
`= 150°`
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