Q. 74.3( 17 Votes )

Two equal circles

Answer :


Property 1: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.


Property 2: Sum of all angles of a straight line = 180°.


Property 3: Sum of all angles of a triangle = 180°.


By property 1, ∆OAB is right-angled at OAB (i.e., OAB = 90°) and PBA is right-angled at PBA (i.e., PBA = 90°)


Clearly,


b + c = OAB


b + c = 90°


b = 90° - c


Similarly,


d + e = PBA


d + e = 90°


e = 90° - d


Now,


a = b = 90° - c [ OA = OC (Radius)]


And,


e = f = 90° - d [ PB = PC (Radius)]


By property 2,


a + f + ACB = 180°


ACB = 180° a f


ACB = 180° (90° - c) (90° - d)


ACB = 180° 90° + c 90° + d


ACB = c + d


Now, in ∆ACB


By property 3,


ACB + c + d = 180°


ACB + ACB = 180° [∵∠ACB = c + d]


2ACB = 180°



ACB = 90°


Hence, ACB = 90°

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Area Related to Circles- Important Formula and ConceptsArea Related to Circles- Important Formula and ConceptsArea Related to Circles- Important Formula and Concepts59 mins
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
caricature
view all courses