PA and PB are two tangents drawn from an external point P to a circle with centre O. If ∠APB = 80°, then ∠POA = __________.

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Solution

PA and PB are tangents drawn from P to circle with centre O.

∠APB = 80º

We know that, the two tangents drawn from an external point to a circle are equally inclined to the segment joining the centre to the point.

∴ ∠APO = ∠BPO = $\frac{80 °}{2}$ = 40º

Also, ∠OAP = 90º (Radius is perpendicular to the tangent at the point of contact)

In ∆OAP,

∠APO + ∠OAP + ∠POA = 180º (Angle sum property of triangle)

⇒ 40º + 90º + ∠POA = 180º

⇒ ∠POA = 180º − 130º = 50º

PA and PB are two tangents drawn from an external point P to a circle with centre O. If ∠APB = 80°, then ∠POA = 50°.

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