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Question
In the figure, ∠DBC = 58°. BD is diameter of the circle.
Calculate:
- ∠BDC
- ∠BEC
- ∠BAC
Solution
∠DBC = 58° ...(Given)
Now, BD is the diameter.
∴ ∠DCB = 90° ...(Angle in a semicircle)
i. In ΔBDC,
∠BDC + 90° + 58° = 180° ...(Sum of the angles of a triangle)
∴ ∠BDC = 180° – (90° + 58°) = 32°
ii. BECD is a cyclic quadrilateral.
∵ ∠BEC + ∠BDC = 180° ...(Opposite angles of a cyclic quadrilateral)
∴ ∠BEC = 180° – ∠BDC
∴ ∠BEC = 180° – 32° = 148°
iii. ∠BAC = ∠BDC = 32° ...(Angles in the same segment of a circle)
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