Question
PA and PB are two tangents from a point P outside the
circle with centre O. If A and B are points on the circle such ∠APB = 128°, then ∠OAB is equal to:Solution
Angle made by the point of contact of the tangent and the centre of the circle is 90°. The Sum of all angles of a quadrilateral is 360°. In a triangle angles opposite to equal sides are equal. Sum of all angles of triangle is 180°. As, PA and PB are tangents ∠OAP = 90° ∠OBP = 90° As, OAPB is a quadrilateral, ∠OAP + ∠APB + ∠PBO + ∠BOA = 360° ⇒ 90° + 128° + 90° + ∠BOA = 360° ⇒ ∠BOA = 360° – 308° ⇒ ∠BOA = 52° As, OA = OB (Radii) In ΔOAB, ∠OAB = ∠OBA (In a triangle angles opposite to equal sides are equal) ∠OAB + ∠OBA + ∠BOA = 180° ⇒ 2 × ∠OAB + 52° = 180° ⇒ ∠OAB = (180° – 52°)/2 ∴ ∠OAB = 64°.
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