Question

8.
If PA and PB are tangents to the circle with
centre O, find x

Answer 1

Join OB.

We know that the radius and tangent are perpendicular at their point of contact.

∴ ∠OBP=∠OAP=90°

Now, In a quadrilateral AOBP

⇒ ∠AOB+∠OBP+∠APB+∠OAP=360°

[ Sum of four angles of a quadrilateral is 360°]

⇒ ∠AOB+90°+60°+90°=360°

⇒ 240°+∠AOB=360°

⇒ ∠AOB=120°

Since OA and OB are the radius of a circle then, △AOB is an isosceles triangle.

⇒ ∠AOB+∠OAB+∠OBA=180°

⇒ 120°+2∠OAB=180°

[ Since, ∠OAB=∠OBA ]

⇒ 2∠OAB=60°

∴ ∠OAB=30°

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