In the figure PA and PB are tangents to the circle. If ∠APO=30°, fi

1.	In the figure PA and PB are tangents to the circle. If ∠APO=30°, find ∠AOB
Answer» ∠APO = 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°

In the figure PA and PB are tangents to the circle. If ∠APO=30°, find ∠AOB

Answer»

∠APO = 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°

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