In the figure, given below, ABCD is a cyclic quadrilateral in which ∠BAD = 75°; ∠ABD = 58°
and ∠ADC = 77°. Find:
1. ∠BDC
2. ∠BCD
3. ∠BCA.
and ∠ADC = 77°. Find:
1. ∠BDC
2. ∠BCD
3. ∠BCA.
Exercise 17 (A) | Q 10 | Page 258
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(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)
∠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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