CBSE Class 10 Mathematics Standard 2023: question 28

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Step-by-step solution

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1. Given: A circle with centre O. PA and PB are tangents from an external point P, touching the circle at A and B.

   To prove: $\angle APB+\angle AOB=180^{\circ }$.

   Figure: Draw the circle with centre O and a point P outside it. Draw tangents PA and PB, and join OA and OB to form the quadrilateral OAPB.

   1 mark

2. In quadrilateral OAPB, the angles add up to $360^{\circ }$:

   $\angle OAP+\angle APB+\angle OBP+\angle AOB=360^{\circ }$

   1 mark

3. $\angle OAP=\angle OBP=90^{\circ }$ [tangent ⟂ radius at the point of contact].

   So $90^{\circ }+\angle APB+90^{\circ }+\angle AOB=360^{\circ }$

   ½ mark

4. $\angle APB+\angle AOB=180^{\circ }$

   Hence the angle between the tangents is supplementary to the angle subtended at the centre by AB.

   ½ mark

Answer: $\angle APB+\angle AOB=180^{\circ }$, i.e., the two angles are supplementary.

Revise the chapter

• Chapter: Circles (notes, formulas, where marks are lost and every past question)