Question : ABCD is a cyclic quadrilateral and BC is the diameter of the circle. If $\angle D B C=29^{\circ}$, then $\angle B A D=$?
Option 1: 129°
Option 2: 119°
Option 3: 111°
Option 4: 122°
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Correct Answer: 119°
Solution : In a circle, the angle subtended by the diameter at any point on the circle is always 90°. $\angle BAC = 90^{\circ}$ Given that $\angle DBC = 29^{\circ}$ So, $\angle BCD = 90^{\circ} - 29^{\circ} = 61^{\circ}$ Also, the property of a cyclic quadrilateral states that the sum of opposite angles of the cyclic quadrilateral is 180°. $\angle BAD + \angle BCD = 180^{\circ}$ $\therefore \angle BAD = 180^{\circ} - \angle BCD = 180^{\circ} - 61^{\circ} = 119^{\circ}$ Hence, the correct answer is 119°.
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