Question

A circle has radius ${\displaystyle \sqrt{2}}$ cm. It is divided into two segments by a chord of length 2 cm. Prove that the angle subtended by the chord at a point in major segment is 45°.

Answer 1

Draw a circle having centre O. Let AB = 2 cm be a chord of a circle. A chord AB is divided by the line OM in two equal segments.

To prove: ∠APB = 45°

Here, AN = NB = 1 cm

And OB = ${\displaystyle \sqrt{2}}$ cm

In ΔONB, OB² = ON² + NB²  ...[Use Pythagoras theorem]

⇒ ${\displaystyle {\left(\sqrt{2}\right)}^2=O{N}^2+{\left(1\right)}^2}$

⇒ ON² = 2 – 1 = 1

⇒ ON = 1 cm  ...[Taking positive square root, because distance is always positive]

Also, ∠ONB = 90°  ...[ON is the perpendicular bisector of the chord AB]

∴ ∠NOB = ∠NBO = 45°

Similarly, ∠AON = 45°

Now, ∠AOB = ∠AON + ∠NOB

= 45° + 45°  

= 90°

We know that, chord subtends an angle to the circle is half the angle subtended by it to the centre.

∴ ${\displaystyle \angle APB=\frac{1}{2}\angle AOB}$

= ${\displaystyle \frac{{90}^{\circ }}{2}}$

= 45° 

Hence proved.