5. In the given figure, AB is the tangent of the circle
with centre O. If ∠DOA = 130°, then find the angle
∠DAB.5. In the given figure, AB is the tangent of the circle
with centre O. If ∠DOA = 130°, then find the angle
∠DAB.
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Step-by-step explanation:
(i) DAB + BCD = 180˚ [Oppo. Angles of a cyclic quadrilateral]
DAB + 130˚ = 180˚
DAB = 180˚ – 130˚
DAB = 50˚
(ii) ADB = 90˚ (angle in semi-circle)
In ADB
DAB + ADB + DBA = 180˚ (angle sum property)
50˚ + 90˚ + DBA = 180˚
DBA = 180˚ – 140˚
DBA = 40˚
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