If two tangents inclined at an angle 600 are drawn to a circle of radius 3 cm. The distance of the chord joining the point of contact of the tangents to the circle is 1.5cm. Then length of each tangent is equal to
A
32√3 cm
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B
6 cm
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C
3 cm
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D
3√3 cm
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Solution
The correct option is D3√3 cm Given−PQ&PRaretangentstothecirclewithcenteO.ONisthedistanceofOfromthechordQR.∠QPR=60OTofindout−PQorPR.Solution−PQ&PRaretangentstothecirclewithcenteO.∴PQ=PR....(i)and∠OQP=∠ORP=90o.....(ii)∴∠OQP+∠ORP+∠P=90o+90o+60o=240o(fromii)⟹∠ROQ=360o−240o=120o.....(iii)(anglesumpropertyofquadrilaterals)NowinΔOQROQ=OR(radiiofthesamecircle)∴ΔOQRisisosceles.⟹∠OQR=∠ORQ∴∠OQR+∠ORQ=2∠OQR.∴2∠OQR+∠ROQ=2∠OQR+120o=180o(fromi)&(anglesumpropertyoftriangles)⟹∠OQR=∠ROQ=30o∴∠PQR=∠OQP−∠OQR=90o−30o=60o⟹∠QRP=180o−60o−60o=60o∴ΔPQRisequilaterali.ePQ=QR....(v)NowONisthedistanceofOfromthechordQR.∴ON⊥QRorQN⟹ΔOQNisrightangledatN.So,byPythagorastheorem,QN=√OQ2−ON2=√32−1.52cm=1.5√3cm⟹QR=2×1.5√3cm=3√3cm.∴From(v)PQ=3√3cmAnsOptionD