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Question


In the figure, BC is the diameter, PQ is the tangent, O is the centre of circle and BAQ=65o.
Find AOC.

A
25o
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B
65o
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C
50o
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D
30o
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Solution

The correct option is C 50o

We know tangent is always perpendicular to the radius at the point of contact.
So, OAQ=90o
OAB+BAQ=90o
OAB+65o=90o
OAB=90o65o
OAB=25o

In OAB
OA=OB (radius of circle)
so, OAB=OBA
OBA=25o

Now, minor arc AC subtending AOC at centre and ABC at point B on the circumference of circle.
So,
AOC=2ABC (angle made by arc at centre is double the angle made at any point on rest part of circumference of circle)

AOC=2×25o
AOC=50o

So, option (c) correct.

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