Question

PA and PB are tangents from an external point P to the circle with centre O. If LOPB :30o

then find LAOB.​

Answer 1

Answer:

∠BPO = 30° …given From P we have two tangents PA and PB We know that if we join point P and centre of circle O then the line PO divides the angle between tangents

⇒ ∠APO = ∠OPB = 30° …(i)

∠OAP = ∠OBP = 90° …radius is perpendicular to tangent …

(ii) Consider quadrilateral OAPB ⇒ ∠OAP + ∠APB + ∠PBO + ∠AOB = 360°…sum of angles of quadrilateral From figure ∠APB = ∠APO + ∠OPB

⇒ ∠OAP + ∠APO + ∠OPB + ∠PBO + ∠AOB = 360° Using (i) and (ii)

⇒ 90° + 30° + 30° + 90° + ∠AOB = 360°

⇒ 240° + ∠AOB = 360°

⇒ ∠AOB = 120°

Hence ∠AOB is 120° ←

Step-by-step explanation: