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Question

A circle has radius 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.

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Solution

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 APB=45
Here, AN=NB=1 cm
And OB=2 cm
In ΔONB, OB2=ON2+NB2 [use Pythagoras theorem]
(2)2=ON2+(1)2
ON2=21=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
[ON =NB =1cm,angles opposite to equal sides are also equal]

Similarly,
AON=45

Now,
AOB=AON+NOB
=45+45=90

We know that, chord subtends an angle on the circumference is half the angle subtended by it on the centre.
APB=12 AOB
APB=902=45
Hence proved.

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