Question

two circles touch each other externally at C and AB is a common tangent to the circles then angle ACB is equal?

Answer 1

$\huge\textbf{Solution}$

Let /_PAC = β and /_PBC = α

Now...

PA = PC

In a ∆CAP

/_PAC = /_ACP = β

similarly PB = PC and /_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°