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A circle has radius sqrt2cm it is divide...

A circle has radius `sqrt2`cm it is divided into 2 segments by a chord of length 2cm prove that angle subtended by the chord at a point in major segment is `45^@`

Text Solution

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Let O be the centre of the given circle and let AB be a chord of length 2 cm.
`:. OA = OB = sqrt(2) cm`.
Clearly, `OA^(2) =+ OB^(2) = {(sqrt(2))^(2) + (sqrt(2))^(2)}cm^(2)`
`= 4cm^(2) = (2cm)^(2) = AB^(2)`.

`:. OA^(2)+OB^(2)=AB^(2)`
Consequently , `/_ AOB = 90^(@)`
Let C be a point in the major segment of the circle . Join CA and CB
Since the angle sbtended by an arc of a circle at its centre is twice the angle subtended by it at any point on the remaining part of the circle , we have
`/_ ACB = (1)/(2) /_ AOB = ((1)/(2) xx 90^(@)) = 45^(@)`
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