In the following figure, XY and X’Y’ are two parallel tangents to

1.	In the following figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.
Answer» Data: XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B To Prove: ∠AOB = 90°. Tangent XY \|\| Tangent X’Y’ ∴ ∠PAB + ∠QBA = 180° (∵ interior angles) \(\frac{1}{2}\)∠PAB + \(\frac{1}{2}\)∠QBA = \(\frac{1}{2}\) x 180° ∠OAB + ∠OBA = 90° Now, in AOAB, ∠AOB + ∠OAB + ∠OBA = 180° ∠AOB + 90° = 180° ∴ ∠AOB = 180° – 90° ∴ ∠AOB = 90°.

In the following figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

Answer»

Data: XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B

To Prove: ∠AOB = 90°.

Tangent XY || Tangent X’Y’

∴ ∠PAB + ∠QBA = 180° (∵ interior angles)

\(\frac{1}{2}\)∠PAB + \(\frac{1}{2}\)∠QBA = \(\frac{1}{2}\) x 180°

∠OAB + ∠OBA = 90°

Now, in AOAB,

∠AOB + ∠OAB + ∠OBA = 180°

∠AOB + 90° = 180°

∴ ∠AOB = 180° – 90°

∴ ∠AOB = 90°.

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In the following figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

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