Radius of a circle is√2cm. This circle is divides into two segment by a chord of length 2 cm.

Question

Radius of a circle is \(\sqrt { 2 }\) cm. This circle is divides into two segment by a chord of length 2 cm. Prove that their chord subtends angle of 45° at a point in major segment.

Answer 1

Given :

O is center of circle, on which minor arc BC subtends ∠BAC is major segment and OB = \(\sqrt { 2 }\) cm BC = 2 cm

To Prove : ∠BAC = 45°

Construction : From center O of circle draw perpendicular to chord BC.

OM ⊥ BC

∴ ∠OMB = ∠OMC = 90°

Proof : We know that ⊥ drawn from center of the circle bisects the chord.

= \(\frac { 1 }{ 2 }\) × 90°

= 45°

Thus, ∠BAC = 45°