CBSE Class 10 Mathematics Standard 2024: question 31

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Step-by-step solution

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1. Figure: draw a circle with centre O and a chord AB. Draw tangents at A and B that meet at an external point P. Join PA and PB.

   Given: AB is a chord, and PA and PB are tangents at A and B.

   To prove: $\angle PAB=\angle PBA$.

   1 mark

2. In $△PAB$: $PA=PB$ [lengths of tangents drawn from an external point to a circle are equal].

   1 mark

3. So $△PAB$ is isosceles. Hence $\angle PAB=\angle PBA$ [angles opposite equal sides are equal]. Therefore the tangents make equal angles with the chord.

   (Alternative: $OA\perp PA$ and $OB\perp PB$ [tangent ⟂ radius], and $\angle OAB=\angle OBA$ since $OA=OB$. So $\angle PAB=90^{\circ }-\angle OAB=90^{\circ }-\angle OBA=\angle PBA$. This version also works when AB is a diameter and the tangents are parallel.)

   1 mark

Answer: Since PA = PB, ∠PAB = ∠PBA; the tangents make equal angles with the chord (proved)

Revise the chapter

• Chapter: Circles (notes, formulas, where marks are lost and every past question)
• Topic: Solving Problems Using Tangent Properties