In the given figure, AOB is a diameter and ABCD is a cyclic quadrilateral.

Question:

In the given figure, AOB is a diameter and ABCD is a cyclic quadrilateral. If ADC = 120°, then ∠BAC = ?
(a) 60°

(b) 30°
(c) 20°
(d) 45°

 

Solution:

(b) 30°
We have:
∠ABC + ∠ADC = 180°     (Opposite angles of a cyclic quadrilateral)
⇒ ∠ABC + 120° = 180°
⇒ ∠ABC = (180° - 120°) = 60°
Also, ∠ACB = 90°     (Angle in a semicircle)
In 
ΔABC, we have:
∠BAC + ∠ACB  + ∠ABC = 180°    (Angle sum property of a triangle)
⇒ ∠BAC + 90° + 60° = 180°
⇒ ∠BAC = (180° - 150°) = 30°

 

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