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प्रश्न
If tangents PA and PB drawn from an external point P to a circle with centre O are inclined to each other at an angle of 80°, then find ∠POA.
योग
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उत्तर
Given that, PA and PB are tangents.

∴ PA and PB are the tangents
∴ PO will be angle bisector of ∠P
Hence, ∠APO = 40°
Now, in ΔAPO,
∠P + ∠APO + ∠POA = 180°
90° + 40° + ∠POA = 180°
130° + ∠POA = 180°
∠POA = 180° – 130°
∠POA = 50°
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