The radius of a circle is This circle is divided into two segments by a chord of length 2 cm. Prove that the angle subtended by this chord at any point of the major segment is 45°.
Given radius of circle OQ = OR = √2 cm
Length of chord, QR = 2 cm
In ΔOQR,
⇒ OQ2 + OR2 = (√2)2 + (√2)2
= 2 + 2
= 4
= (QR)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°.
Hence proved