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Question

In the given figure, XY and X1Y1 are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X1Y1 at B then AOB is ___.

A
100
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B
60
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C
80
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D
90
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Solution

The correct option is D 90

In the figure, if we join point 'O' and 'C'. Then we can see that OC = DA = EB,
OE = CB and OD = CA
OC = OD = OE [radius of circle]
ODAC and OEBC are squares

By Theorem - The centre of a circle lies on the bisector of the angle between two tangents drawn from a point outside it.

AO bisects CAD and BO bisects CBE.
This means AO and BO are diagonals of square ODAC and OBEC respectively.
So, AOC = 45 and BOC = 45.

Now, in AOB
AOB = AOC + BOC
AOB = 45 + 45 = 90

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