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Question
In the figure, given below, ABCD is a cyclic quadrilateral in which ∠BAD = 75°; ∠ABD = 58° and ∠ADC = 77°. Find:
- ∠BDC,
- ∠BCD,
- ∠BCA.
Solution
i. By angle – sum property of triangle ABD,
∠BAD + ∠ABD + ∠ADB = 180°
133° + ∠ADB = 180°
∠ADB = 180° – 133°
∠ADB = 47°
∴ ∠ADC = ∠ADB + ∠BDC
∴ 77° = 47° + ∠BDC
∴ 77° – 47° = ∠BDC
∴ ∠BDC = 30°
ii. ∠BAD + ∠BCD = 180° ...(Sum of opposite angles of a cyclic quadrilateral is 180°)
`=>` ∠BCD = 180° – 75° = 105°
∴ ∠BCD = 105°
iii. ∠BCA = ∠BDA = 47° ...(Angle subtended by the same chord on the circle are equal)
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