Mathematics
In the figure (i) given below, O is the center of the circle. If ∠COD = 40° and ∠CBE = 100°, then find :
(i) ∠ADC
(ii) ∠DAC
(iii) ∠ODA
(iv) ∠OCA
Answer
(i) ABCD is a cyclic quadrilateral.
∵ exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.
∠ADC = ∠CBE = 100°.
Hence, the value of ∠ADC = 100°
(ii) Arc DC subtends ∠DOC at center and ∠DAC at point A.
⇒ ∠DOC = 2∠DAC (∵ angle subtended on center by an arc is double the angle subtended on the remaining part of circle.)
⇒ 40° = 2∠DAC
⇒ ∠DAC = 20°
Hence, the value of ∠DAC = 20°.
(iii) In △COD, OC = OD = radius of the same circle.
∠CDO = ∠DCO = x
Since sum of angles of triangle = 180°
In △COD
⇒ ∠CDO + ∠DCO + ∠COD = 180°
⇒ x + x + 40° = 180°
⇒ 40° + 2x = 180°
⇒ 2x = 180° - 40°
⇒ 2x = 140°
⇒ x = 70°.
From figure,
⇒ ∠ADC = ∠ODA + ∠CDO
⇒ 100° = ∠ODA + 70°
⇒ ∠ODA = 100° - 70° = 30°.
Hence, the value of ∠ODA = 30°.
(iv) Since sum of angles of triangle = 180°
In △ADC
⇒ ∠ADC + ∠DAC + ∠ACD = 180°
⇒ 100° + 20° + ∠ACD = 180°
⇒ 120° + ∠ACD = 180°
⇒ ∠ACD = 180° - 120°
⇒ ∠ACD = 60°.
From figure,
∠OCA = ∠DCO - ∠ACD = 70° - 60° = 10°.
Hence, the value of ∠OCA = 10°.
Related Questions
In the figure (i) given below, PQRS is a cyclic quadrilateral in which PQ = QR and RS is produced to T. If ∠QPR = 52°, calculate ∠PST.
In the figure (ii) given below, O is the center of the circle. If ∠BAD = 75° and BC = CD, find:
(i) ∠BOD
(ii) ∠BCD
(iii) ∠BOC
(iv) ∠OBD
In the adjoining figure, O is the center and AOE is the diameter of the semicircle ABCDE. If AB = BC and ∠AEB = 50°, find :
(i) ∠CBE
(ii) ∠CDE
(iii) ∠AOB.
Prove that OB is parallel to EC.
In the figure (ii) given below, O is the center of the circle. If ∠OAD = 50°, find the values of x and y.
