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12. Two tangents PA and PB are drawn to a cirelewith centre O from an external point P. Prove that[2016LAPB2ZOAB |
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Answer» Let ∠APB= x° We know that the tangents to a circle from an external point are equal in length so PA= PB.PA =PB ∠PBA = ∠PAB[Angles opposite to the equal sides of a triangle are equal.] ∠APB+ ∠PBA +∠PAB= 180°[Sum of the angles of a triangle is 180°] x° + ∠PAB +∠PAB = 180° [∠PBA = ∠PAB] x° + 2∠PAB = 180° ∠PAB =½(180° - x°) ∠PAB =90° - x°/2∠OAB +∠PAB=90°∠OAB =90° - ∠PAB∠OAB =90° - (90° - x°/2)∠OAB =90° - 90° + x°/2∠OAB = x°/2∠OAB = ∠APB /2∠OAB = 1/2∠APB ∠APB = 2∠OAB |
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