Answer :

Given radius of circle OQ = OR = √2 cm

Length of chord, QR = 2 cm

In ΔOQR,

⇒ OQ^{2} + OR^{2} = (√2)^{2} + (√2)^{2}

= 2 + 2

= 4

= (QR)^{2}

⇒ OQ^{2} + OR^{2} = QR^{2}

∴ ΔOQR is a right angled triangle at ∠QOR.

We know that angle at the centre is double the angle on the remaining part of the circle.

⇒∠QOR = 2∠QPR

⇒ 90° = 2∠QPR

∴ ∠QPR = 45°

∴ The angle subtended by the chord at the point in major segment is 45°.

Hence proved

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