Given radius of circle OQ = OR = √2 cm
Length of chord, QR = 2 cm
⇒ OQ2 + OR2 = (√2)2 + (√2)2
= 2 + 2
⇒ OQ2 + OR2 = QR2
∴ 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°.
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