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प्रश्न
In the given figure, AB is the diameter of a circle with centre O. ∠BCD = 130°. Find:
(i) ∠DAB
(ii) ∠DBA
उत्तर
i. ABCD is a cyclic quadrilateral
m∠DAB = 180° − ∠DCB
= 180° - 130°
= 50°
ii. In ∆ADB,
m∠DAB + m∠ADB + m∠DBA = 180°
⇒ 50° + 90° + m∠DBA = 180°
⇒ m∠DBA = 40°
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