Question:
In figure, BC is a diameter of the circle and ∠BAO = 60°. Then, ∠ADC is equal to OA = OB [both are the radius of circle]
(a) $30^{\circ}$
(d) $45^{\circ}$
(d) $60^{\circ}$
(d) $120^{\circ}$
Solution:
(c) In ΔAOB,
∠OBA = ∠BAO
[angles opposite to equal sides are equal]
∠OBA = 60° [∴ ∠BAO = 60°, given]
∠ABC=∠ADC
[angles in the same segment AC are equal]
∠ADC = 60°
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