Prove that the angle between the two tangents drawn from an external p

1.	Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Answer» To Prove: ∠APB + ∠AOB = 180° OA is radius, PA is tangent. ∴∠PAO = 90° OB is radius, PB is tangent. ∴ ∠PBO = 90° Now OAPB is a quadrilateral. ∴∠PAO + ∠PBO = 90° + 90° = 180° Sum of four angles of a quadrilateral is 360° ∴ ∠PAO + ∠PBO + ∠APB + ∠AOB = 360° 180° + ∠APB + ∠AOB = 360° ∠APB + ∠AOB = 360° – 180° ∴ ∠APB + ∠AOB = 180° If sum of two angles is equal to 180°, then they are supplementary angles. ∴ ∠APB and ∠AOB are supplementary angles.

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Answer»

To Prove: ∠APB + ∠AOB = 180°

OA is radius, PA is tangent.

∴∠PAO = 90° OB is radius, PB is tangent.

∴ ∠PBO = 90°

Now OAPB is a quadrilateral.

∴∠PAO + ∠PBO = 90° + 90° = 180°

Sum of four angles of a quadrilateral is 360°

∴ ∠PAO + ∠PBO + ∠APB + ∠AOB = 360°

180° + ∠APB + ∠AOB = 360°

∠APB + ∠AOB = 360° – 180°

∴ ∠APB + ∠AOB = 180°

If sum of two angles is equal to 180°, then they are supplementary angles.

∴ ∠APB and ∠AOB are supplementary angles.

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Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

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