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°
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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