Answer is C. 30°
Given:
APB = 30°
Property 1: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.
Property 2: Sum of all angles of a triangle = 180°
By property 1,
PA = PB (tangent from P)
And,
∠PAB = ∠PBA [∵PA = PB]
By property 2,
∠PAB + ∠PBA + ∠APB = 180°
⇒ ∠PAB + ∠PBA + 30° = 180°
⇒ ∠PAB + ∠PBA = 180° - 30°
⇒ ∠ PAB + ∠ PBA = 150°
⇒ ∠ PBA + ∠ PBA = 150° [∵∠PAB = ∠PBA]
⇒ 2∠PBA = 150°
⇒ ∠PBA = \(\frac{150^°}{2}\)
⇒ ∠PBA = 75°
Now,
∠PBA = ∠CAB = 75° [Alternate angles]
∠PBA = ∠ACB = 75° [Alternate segment theorem]
Again by property 2,
∠CAB + ∠ACB + ∠CBA = 180°
⇒ 75° + 75° + ∠CBA = 180°
⇒ 150° + ∠CBA = 180°
⇒ ∠CBA = 180° - 150°
⇒ ∠CBA = 30°
Hence, ∠CBA = 30°