Question

TWO CIRCLES TOUCH EACH OTHER EXTERNALLY AT 'C' AND 'AB' IS COMMON TANGENT TO CIRCLES. THEN ANGLE 'ACB' =


60

45

30

90


AND PLEASE TELL HOW TO SOLVE

Answer 1

this is your answer hope this will help you

Answer 2

$\huge\textbf{Solution}$

Let /_PAC = β and /_PBC = α

Now...

PA = PC

In a ∆CAP

/_PAC = /_ACP = β

similarly PB = PC

/_PCB = /_CBP = α

Now

In the ∆ACB,

/_CAB + /_CBA + /_ACB = 180°  

As sum of the interior angles in a triangle is 180°

β + α + (β + α) = 180°