two circle touch each other externally at c and ab is a common tangent to the cirle. then ,angleACB =?
Let A be on a circle woth centre O and B be the point on the circle with O' as centre. And AB be the tangent to both circles touching at A and B.
Let the two circles touch at C.
Let the tangent at C meet AB at N.
Now NA and NT are tangents to the the circle with centre O andtherefore NA= NB. Sothetriangle NAC is isosceles and angles NAC = NCA = x say.
By similar consideration NB and NT are tangents from N to circle with centre O'. So triangle NBC is isosceles with NB=NC and therefore, angles NBC = NCB = y say.
Therefore in triangle ABC, angles A+B+C = x + y + (x+y) = 180
Or 2(x+y) =180.
x+y = 180/2 = 90.
x+y = angle ACB =180/2 =90 degree = angle ACB