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Question
In Fig. 1, PA and PB are tangents to the circle with centre O such that ∠APB = 50°. Write the measure of ∠OAB.
Solution
PA and PB are tangents drawn from an external point P to the circle.
∴ PA = PB (Length of tangents drawn from an external point to the circle are equal.)
In ∆PAB,
PA = PB
⇒ ∠PBA = ∠PAB .....(1) (Angles opposite to equal sides are equal.)
Now,
∠APB + ∠PBA + ∠PAB = 180°
⇒ 50º + ∠PAB + ∠PAB = 180° [Using (1)]
⇒ 2∠PAB = 130°
⇒ ∠PAB =`130^@/2`= 65°
We know that radius is perpendicular to the tangent at the point of contact.
∴ ∠OAP = 90° (OA ⊥ PA)
⇒ ∠PAB + ∠OAB = 90°
⇒ 65° + ∠OAB = 90°
⇒∠OAB = 90° − 65° = 25°
Hence, the measure of ∠OAB is 25°.
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