From a point P two tangents PA and PB are drawn such that PA 9 cm and angle APB = 60 then AB is

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Given: PA and PB are tangents of a circle, PA = 5 cm and ∠APB = 60° Let O be the center of the given circle and C be the point of intersection of OP and AB In ΔPAC and ΔPBC (adsbygoogle = window.adsbygoogle || []).push({}); PA = PB (Tangents from an external point are equal) ∠APC = ∠BPC (Tangents from an external point are equally inclined to the segment joining center to that point) PC = PC (Common) Thus ΔPACΔPBC (By SAS congruency rule) ..........(1) ∴ AC = BC Also ∠APB = ∠APC + ∠BPC AB = AC + BC = AC + AC (AC = BC) ⇒ AB = MathematicsSecondary School Mathematics XStandard X (adsbygoogle = window.adsbygoogle || []).push({});

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PA and PB are the tangents drawn to a circle with centre O. If PA= 6 cm and  APB = 600 , then find the length of the chord AB. We know that Tangent from external point are equal ∴ PA = PB Since angles opposite equal sides are equal Thus, ∠ PAB = ∠ PBA By Angle Sum property in Δ APB ∠ APB + ∠ PAB + ∠ PBA = 180° 60° + ∠ PAB + ∠ PAB = 180° 60° + 2∠ PAB = 180° 2∠ PAB = 180° − 60° 2∠ PAB = 120° ∠ PAB = (120° )/2 ∠ PAB = 60° Thus, in Δ APB ∠ PAB = ∠ PBA = 60° And, ∠ APB = 60° Since all angles are 60°, ∴ Δ APB is an equilateral triangle 60° + ∠ PAB + ∠ PAB = 180° 60° + 2∠ PAB = 180° 2∠ PAB = 180° − 60° 2∠ PAB = 120° ∠ PAB = (120° )/2 ∠ PAB = 60° Thus, in Δ APB ∠ PAB = ∠ PBA = 60° And, ∠ APB = 60° Since all angles are 60°, ∴ Δ APB is an equilateral triangle Since in equilateral triangle All sides are equal ∴ AB = AP = 6 cm

In the given figure, PA and PB are tangents to the given circle such that PA = 5 and angle APB = 60o The length of chord AB is a 5√2 cm b 5 cm a 5√3 cm a 7.5 cm

Solution

Given: PA and PB are tangents of a circle, PA = 5 cm and ∠APB = 60° Let O be the center of the given circle and C be the point of intersection of OP and AB In ΔPAC and ΔPBC (adsbygoogle = window.adsbygoogle || []).push({}); PA = PB (Tangents from an external point are equal) ∠APC = ∠BPC (Tangents from an external point are equally inclined to the segment joining center to that point) PC = PC (Common) Thus ΔPACΔPBC (By SAS congruency rule) ..........(1) ∴ AC = BC Also ∠APB = ∠APC + ∠BPC AB = AC + BC = AC + AC (AC = BC) ⇒ AB = MathematicsSecondary School Mathematics XStandard X (adsbygoogle = window.adsbygoogle || []).push({});

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