# In Fig. 10.89, if

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°

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses