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Question
In the given figure, AB is the diameter of a circle with centre O. ∠BCD = 130o. Find:
1) ∠DAB
2) ∠DBA
Solution
Given: AB is the diameter of the circle with centre O, ∠BCD = 130°
To find: ∠DAB, ∠DBA
1) Clearly, ABCD is a cyclic quadrilateral.
We know, the sum of a pair of opposite angles of a cyclic quadrilateral is 180°
∴ ∠DAB + ∠DCB = 180°
⇒ ∠DAB + 130° = 180°
⇒ ∠DAB = 180° - 130° = 50°
2) Consider ΔDAB,
Here, ∠ADB = 90° …..[Since angle in a semi-circle is a right angle]
So, by angle sum property of a triangle,
∠DAB + ∠DBA + ∠ADB = 180°
⇒ 50° + ∠DBA + 90° = 180°
⇒ ∠DBA = 180° - 140° = 40°
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