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for a cirlce with centre O , two tangent PA and PB from an angle of 80° from an external point P, find |
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Answer» I ANSWER is the 50Step-by-step EXPLANATION:→OA,OB are radii,PA,PB are tangent→∆PAO=90°,∆PBO=90°,∆BPA=80°In quadrilateral OAPB,→90°+90°+80°+∆AOB=360°→960°-180°-80°=100°But ∆POA=1/2∆AOB→1/2×100=50°=∆POA=50° |
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