Question

A circle has radius $\sqrt{2}$ 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 the major segment is ${45}^{\circ }$.

Answer 1

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


To prove that $\angle APB={45}^{\circ }$
Here, $AN=NB=1cm$
And $OB=\sqrt{2}cm$
$In\Delta ONB,O{B}^2=O{N}^2+N{B}^2$ [use Pythagoras theorem]
$\Rightarrow (\sqrt{2}{)}^2=O{N}^2+(1{)}^2$
$\Rightarrow O{N}^2=2-1=1$
$ON=1cm$
[taking positive square root, because distance is always positive]

Also, $\angle ONB={90}^{\circ }$
[ON is the perpendicular bisector of the chord AB]

$\therefore \angle NOB=\angle NBO={45}^{\circ }$
[ON =NB =1cm,angles opposite to equal sides are also equal]

Similarly,
$\angle AON={45}^{\circ }$

Now,
$\angle AOB=\angle AON+\angle NOB$
$={45}^{\circ }+{45}^{\circ }={90}^{\circ }$

We know that, chord subtends an angle on the circumference is half the angle subtended by it on the centre.
$\therefore \angle APB=\frac{1}{2}\angle AOB$
$\angle APB=\frac{{90}^{\circ }}{2}={45}^{\circ }$
Hence proved.