Answer :
(i)
ABCD is a cyclic quadrilateral.
∴ ∠BAD + ∠BCD = 180°
⇒ ∠BAD + 120° = 180°
⇒ ∠BAD = 60°
(ii)
∠BDA = 90°[Angle in a semi-circle]
In triangle ABD,
∠ABD + ∠BDA + ∠BAD = 180°[Sum of angles of triangle]
⇒ ∠ABD + 90° + 60° = 180°
⇒ ∠ABD + 150° = 180°
⇒ ∠ABD = 30°
(iii)
OD = OA [Radius]
∴ ∠OAD = ∠ODA = ∠BAD = 180°
∴∠ODB = 90° - ∠ODA
⇒ ∠ODB = 90° - 60°
⇒ ∠ODB = 30°
(iv)
∠ADC = ∠ADB + ∠CDB
⇒ ∠ADC = 90° + 30°
⇒ ∠ADC = 120°
In triangle AOD,
∠AOD + ∠OAD + ∠ODA = 180°[Sum of angles of triangle]
⇒ ∠AOD + 60° + 60° = 180°
⇒ ∠AOD + 120° = 180°
⇒ ∠AOD = 60°
∴ Triangle AOD is an equilateral triangle.
Rate this question :






















In Fig. 16.195, <
RD Sharma - Mathematics<span lang="EN-US
RS Aggarwal & V Aggarwal - Mathematics<span lang="EN-US
RS Aggarwal & V Aggarwal - MathematicsPQRS is a
RD Sharma - MathematicsABCD is a
RD Sharma - Mathematics<span lang="EN-US
RS Aggarwal & V Aggarwal - Mathematics<span lang="EN-US
RS Aggarwal & V Aggarwal - Mathematics<span lang="EN-US
RS Aggarwal & V Aggarwal - Mathematics<span lang="EN-US
RS Aggarwal & V Aggarwal - Mathematics<span lang="EN-US
RS Aggarwal & V Aggarwal - Mathematics