Question

A circle with centre 'O' has a radius $\sqrt{2}$ cm, it is divided into two segments by a chord AB of length 2 cm. Prove that the angle subtended by the chord at a point 'P' in the major segment is ${45}^0$.

Answer 1

Given radius $=\sqrt{2}cm$

Therefore $AO=\sqrt{2}cm$

Let OD be the perpendicular from O on AB

And AB $=2cm$

Therefore $AD=1cm$ (perpendicular from the centre bisects the chord)

Now in triangle AOD,

$AO=\sqrt{2}cm$

$AD=1cm$

And let angle AOD $=\theta$

Therefore , $sin\theta =\frac{1}{\sqrt{2}}$

Hence, $\theta ={45}^o$

Therefore angle AOB $={45}^o+{45}^o={90}^o$

Then angle APB $=\frac{90}{2}={45}^o$ {angle made by a chord at the centre is double of the angle made by the chord at any poin on the circumference)