Q17 of 141 Page 447

In the given figure, BOC is a diameter of a circle with centre O. If ∠BCA = 30°, then ∠CDA = ?

Given:


Here,


∠BAC = 90° (angle in the semicircle)


Now, in ΔABC


By angle sum property


∠BCA + ∠BAC + ∠ABC = 180°


30° + 90° + ∠ABC = 180°


∠ABC = 180° – 30° – 90°


∠ABC = 60°


Here,


∠ABC = ∠ADC (angles in the same segment)


∴ ∠CDA = 60°

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