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In the Given Figure, Ab is a Diameter of the Circle with Centre O. Do is Parallel to Cb and ∠Dcb = 120°. Calculate: (I) ∠Dab, (Ii) ∠Dba, (Iii) ∠Dbc, (Iv) ∠Adc. Also Show that the δAod is an Equilatera - ICSE Class 10 - Mathematics

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Question

In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate:
(i) ∠DAB, (ii) ∠DBA, (iii) ∠DBC, (iv) ∠ADC.
Also show that the ΔAOD is an equilateral triangle .

Solution


(i) ABCD is a cyclic quadrilateral
∴ ∠DCB + ∠DAB = 180°
(pair of opposite angles in a cyclic quadrilateral are supplementary)
⇒ ∠DAB =180° -120° = 60°

(ii) ∠ADB = 90°
(Angle in a semicircle is a right angle)
 ∴ ∠DBA = 90° - ∠DAB = 90° - 60°= 30°

(iii) OD = OB
∴ ∠ODB = ∠OBD
Or ∠ABD = 30°
Also, AB || ED
∴∠DBC = ∠ODB = 30°  (Alternate angles)

(iv) ∠ABD + ∠DBC = 30°  + 30° = 60° 
⇒ ∠ABC = 60°
In cyclic quadrilateral ABCD,
∠ADC + ∠ABC =180°
(pair of opposite angles in a cyclic quadrilateral are supplementary)
⇒ ∠ADC =180° - 60° =120°
In ∆AOD, OA = OD (radii of the same circle)

∠AOD =∠ DAO Or ∠DAB = 60° [proved in (i)]

∠DAO = 60°

⇒ ∠ADO = ∠AOD =∠DAO = 60°
∴ ∆AOD is an equilateral triangle.

  Is there an error in this question or solution?

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Solution In the Given Figure, Ab is a Diameter of the Circle with Centre O. Do is Parallel to Cb and ∠Dcb = 120°. Calculate: (I) ∠Dab, (Ii) ∠Dba, (Iii) ∠Dbc, (Iv) ∠Adc. Also Show that the δAod is an Equilatera Concept: Arc and Chord Properties - Angle in a Semi-circle is a Right Angle.
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