Q. 275.0( 5 Votes )

In Fig. 10.89, if

Answer :

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]


2PBA = 150°



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°

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