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प्रश्न
In the given figure, AB is the diameter of a circle with centre O.
∠BCD = 130°. Find:
- ∠DAB
- ∠DBA
उत्तर
i. ∠DAB + ∠BCD = 180° ...(Opposite angle of a cyclic quadrilateral)
∠DAB + 130° = 180°
∠DAB = 180° – 130°
∠DAB = 50°
ii. ∠ADB = 90° ...(Angle in semicircle)
In ΔADB,
∠DAB + ∠ADB + ∠DBA = 180° ...(Angle sum property)
50° + 90° + ∠DBA = 180°
∠DBA = 180° – 140°
∠DBA = 40°
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संबंधित प्रश्न
In the given figure, AB is the diameter of a circle with centre O. ∠BCD = 130o. Find:
1) ∠DAB
2) ∠DBA
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