Question

If tangents PA and PB from a point P to a circle with centre I are inclined to each other at angle 80 find POA
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Answer 1

OA and OB are radii of the circle to the tangents PA and PB respectively. ∴ OA ⊥ PA and, ∴ OB ⊥ PB ∠OBP = ∠OAP = 90° In quadrilateral AOBP, Sum of all interior angles = 360° ∠AOB + ∠OBP + ∠OAP + ∠APB  = 360° ⇒ ∠AOB + 90° + 90° + 80°  = 360° ⇒ ∠AOB = 100° Now, In ΔOPB and ΔOPA, AP = BP (Tangents from a point are equal) OA = OB (Radii of the circle) OP = OP (Common side) ∴ ΔOPB ≅ ΔOPA (by SSS congruence condition) Thus ∠POB = ∠POA ∠AOB = ∠POB + ∠POA ⇒ 2 ∠POA = ∠AOB ⇒ ∠POA = 100°/2 = 50° ∠POA is equal to option  (A) 50° Read more on Sarthaks.com - https://www.sarthaks.com/2661/tangents-and-from-point-circle-with-centre-are-inclined-each-other-angle-80-then-poa-equal