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°