Two tangents PA and PB are drawn from an external point P to a circle with centre O, such that ∠𝐴𝑃𝐵 = ∠𝑥 and ∠𝐴𝑂𝐵 = 𝑦. Prove that opposite angles are supplementary.

It is given that Two tangents 𝑃𝐴 and 𝑃𝐵 are drawn from an external point 𝑃 to a circle with centre O, such that ∠𝐴𝑃𝐵and ∠𝐴𝑂𝐵 = 𝑦 

We know that the tangent at a point to a circle is perpendicular to radius through that point. Therefore 𝑂𝐴⟂𝐴𝑃 and 𝑂𝐵⟂𝐵𝑃 

So ∠𝑂𝐴𝑃 = 90° 

⇒ ∠𝑂𝐵𝑃 = 90° 

⇒ ∠𝑂𝐴𝑃 + ∠𝑂𝐵𝑃 = 90° + 90°…….. (𝑖) 

⇒ = 180°

In a quadrilateral 𝐴𝑂𝐵𝑃 the sum of all angles = 360° 

So ∠𝑂𝐴𝑃 + ∠𝑂𝐵𝑃 + ∠𝐴𝑃𝐵∠ + ∠𝐴𝑂𝐵 = 360° 

⇒ 180° + ∠𝐴𝑃𝐵 + ∠𝐴𝑂𝐵 = 360° [𝑓𝑟𝑜𝑚 (𝑖)] 

⇒ ∠𝐴𝑃𝐵 + ∠𝐴𝑂𝐵 = 360° − 180° 

⇒ ∠𝐴𝑃𝐵 + ∠𝐴𝑂𝐵 = 180° 

Hence, the sum of opposite angles are supplementary.