Question

In the given figure, AB is the tangent of the circle

with centre O. If ∠DOA = 130°, then find the angle

∠DAB.​

Answer 1

Step-by-step explanation:

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˚

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DBA = 40˚