PA and PB are tangents from an external point P to a circle with centre O. Prove that ∠APB + ∠AOB = 180°.

How to lay out the answer. Follow the frame, then write it in your own words.

1 · Write what is given and what is to be found Given: tangents PA and PB from P to a circle with centre O. To prove: ∠APB + ∠AOB = 180°.
2 · Draw the figure Circle with centre O; tangents PA and PB; join OA and OB; mark right angles at A and B.
3 · Write the formula or theorem The tangent is perpendicular to the radius: ∠OAP = ∠OBP = 90°.
4 · Use the angle sum In quadrilateral OAPB: ∠OAP + ∠APB + ∠PBO + ∠BOA = 360°.
5 · Substitute 90° + ∠APB + 90° + ∠AOB = 360°, so ∠APB + ∠AOB = 180°.
6 · Check and box the answer The two angles are supplementary. Write 'Hence proved'.

Scoring tip: Name the quadrilateral OAPB and write its angle sum, 360°, in words. That is the step most students skip.