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In figure, XY and X are two parallel tangents to a circle with centre 0 and another tangent AB with point of contact C intersecting XY at A and X at B. then is
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(b): Since, tangents drawn from an external point to a circle are equal. AP = AC Thus, in and AP = AC AO = AO (Common side) OP = OC (Radius of circle) By SSS criterion of congruence, we have ⇒. Similarly, we can prove that Since, XY ∠PAC+∠QBC=180 (Sum of interior angles on the same side of transversal is∘∴2 ∠CAO+2 ∠AOB=180 .... (i) ∠CAO+∠CBO=90 In ⇒∘⇒∘ ... (ii) From Eq. (i) and (ii), we get ∘∘∴∠AOB=90\displaystyle∘
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Question Text | In figure, XY and X are two parallel tangents to a circle with centre 0 and another tangent AB with point of contact C intersecting XY at A and X at B. then is |
Answer Type | Text solution:1 |
Upvotes | 150 |