ABCD is a cyclic quadrilateral in which:
(i) BC||AD, ∠ADC=110° and ∠BAC=50°. Find ∠DAC.
(ii) ∠DBC=80° and ∠BAC=40° Find ∠BCD.
(iii) ∠BCD=100° and ∠ABD=70°. Find ∠ADB.
Answers
Given : ABCD is a cyclic quadrilateral in which:
(i) BC || AD, ∠ADC = 110° and ∠BAC = 50°
(ii) ∠DBC = 80° and ∠BAC = 40°
(iii) ∠BCD = 100° and ∠ABD = 70°.
To Find : (i) ∠DAC.
(ii) ∠BCD.
(iii) ∠ADB.
Proof :
(i) Since, ABCD is a cyclic quadrilateral, and Sum of Opposite pair of angles in a cyclic quadrilateral is 180° :
∴ ∠ABC + ∠ADC = 180°
∠ABC + 110° = 180°
∠ABC = 180° - 110°
∠ABC = 70°
Since, AD ‖ BC and sum of the interior angles on the same side of a transversal is 180°.
∴ ∠DAB + ∠ABC = 180°
∠DAC + 50° + 70° = 180°
∠DAC + 120° = 180°
∠DAC = 180° - 120°
∠DAC = 60°
Hence, ∠DAC is 60°.
(ii) Since angles in the same segment of a circle are equal :
∴ ∠BAC = ∠BDC = 40°
In ∆BDC ,
Since Sum of the angles of a triangle is 180° :
∴∠DBC + ∠BCD + ∠BDC = 180°
80° + ∠BCD + 40° = 180°
120° + ∠BCD = 180°
∠BCD = 180° - 120°
∠BCD = 60°
Hence , ∠BCD is 60°.
(iii) Since, ABCD is a cyclic quadrilateral, and Sum of Opposite pair of angles in a cyclic quadrilateral is 180° :
∴ ∠BAD + ∠BCD = 180°
∠BAD + 100° = 180°
[Given :∠BCD = 100°]
∠BAD = 180° - 100°
∠BAD = 80°
In ∆ ABD,
Since Sum of the angles of a triangle is 180° :
∠ABD + ∠ADB + ∠BAD = 180°
70° + ∠ADB + 80° = 180°
[Given : ∠ABD = 70°]
150° + ∠ADB = 180°
∠ADB = 180° - 150°
∠ADB = 30°
Hence, ∠ADB is 30°.
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Answer:-
Given :
ABCD is a cyclic quadrilateral in which:
(i) BC || AD, ∠ADC = 110° and ∠BAC = 50°
(ii) ∠DBC = 80° and ∠BAC = 40°
(iii) ∠BCD = 100° and ∠ABD = 70°.
To Find :
(i) ∠DAC.
(ii) ∠BCD.
(iii) ∠ADB.
Proof :
(i) Since, ABCD is a cyclic quadrilateral, and Sum of Opposite pair of angles in a cyclic quadrilateral is 180° :
∴ ∠ABC + ∠ADC = 180°
∠ABC + 110° = 180°
∠ABC = 180° - 110°
∠ABC = 70°
Since, AD ‖ BC and sum of the interior angles on the same side of a transversal is 180°.
∴ ∠DAB + ∠ABC = 180°
∠DAC + 50° + 70° = 180°
∠DAC + 120° = 180°
∠DAC = 180° - 120°
∠DAC = 60°
Hence, ∠DAC is 60°.
(ii) Since angles in the same segment of a circle are equal :
∴ ∠BAC = ∠BDC = 40°
In ∆BDC ,
Since Sum of the angles of a triangle is 180° :
∴∠DBC + ∠BCD + ∠BDC = 180°
80° + ∠BCD + 40° = 180°
120° + ∠BCD = 180°
∠BCD = 180° - 120°
∠BCD = 60°
Hence , ∠BCD is 60°.
(iii) Since, ABCD is a cyclic quadrilateral, and Sum of Opposite pair of angles in a cyclic quadrilateral is 180° :
∴ ∠BAD + ∠BCD = 180°
∠BAD + 100° = 180°
[Given :∠BCD = 100°]
∠BAD = 180° - 100°
∠BAD = 80°
In ∆ ABD,
Since Sum of the angles of a triangle is 180° :
∠ABD + ∠ADB + ∠BAD = 180°
70° + ∠ADB + 80° = 180°
[Given : ∠ABD = 70°]
150° + ∠ADB = 180°
∠ADB = 180° - 150°
∠ADB = 30°
Hence, ∠ADB is 30°.
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