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Q.

O is the center of a circle. PA and PB are tangents to the circle from a point P. Find the sum of the opposite angles of the quadrilateral PAOB and write what that concludes.

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a

Sum of the opposite angles of the quadrilateral PAOB is

90 o

and hence, it is a cyclic quadrilateral.

b

Sum of the opposite angles of the quadrilateral PAOB is

100 o

and hence, it is a cyclic quadrilateral.

c

Sum of the opposite angles of the quadrilateral PAOB is

80 o

and hence, it is a cyclic quadrilateral.

d

Sum of the opposite angles of the quadrilateral PAOB is

180 o

and hence, it is a cyclic quadrilateral.

answer is A.

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Detailed Solution

Given that O is the center of a circle. PA and PB are tangents to the circle from a point P.
We have to find the sum of the opposite angles of the quadrilateral PAOB.
The tangent of a circle at any point in a circle is perpendicular to the radius through the point of contact.
Consider the figure,
Question Image

Question Image

As we know that the tangent of a circle at any point in a circle is perpendicular to the radius through the point of contact. So,

∠ O B P = ∠ O A P = 90 °

In the quadrilateral OABP,

∠ O B P + ∠ O A P + ∠ A P B + ∠ A O B = 360 ° ⇒ 90 ° + 90 ° + ∠ A P B + ∠ A O B = 360 ° ⇒ ∠ A P B + ∠ A O B = 360 ° − 180 ° ⇒ ∠ A P B + ∠ A O B = 180 °

Here, the sum of the opposite angles of the quadrilateral is

1800

.
Hence, AOBP is a cyclic quadrilateral.
Therefore, the correct option is 1.

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