Question

The radius of a circle is v2 cm. A chord 2 cm in length divides the circle into two
segments. Prove that the angle of larger segment is:
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer 1

Answer:

Given radius =

2

cm

Therefore AO=

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=

2

cm

AD=1cm

And let angle AOD =θ

Therefore , sinθ=

2

1

Hence, θ=45

o

Therefore angle AOB =45

o

+45

o

=90

o

Then angle APB =

2

90

=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)