6 Circle in Equilateral Triangle i have one doubt in this when there are 6 circles in equilateral triangle then how the angle is 30 degree
 A: *

*In the diagram, both the triangles have one angle as 90 degree and two sides as same [1. radius of the triangle 2. common side]. So , both the triangles are congruent. [(The RHS rule: Right-angled, Hypotenuse, Side)]


*

*Hence the angle of triangle will be divided as 60/2 = 30 degree.



Diagram
A: Yes, it is $30^\circ$.  The line through the bottom left corner and the center of the bottom left circle is an angle bisector of the triangle's $60^\circ$ corner.  It must be an angle bisector since we know that at the point in the circle, the line is equidistant from both sides of the angle with distance $r$.  
A: Here's a more Geometric approach:
$OY$ is the radius of circle $O$ and is tangent to $\triangle ABC$.
From that, we have $\triangle YOC \cong\triangle XOC$ because $OY\cong XO$ and $CO\cong CO$. (HL congruency for right triangles)
Therefore, $\angle OCY\cong\angle OCX$. And since they add up to $60^{\circ}$, we have $$\angle OCY+\angle OCX=60\\2\angle OCY=60\\\angle OCY=\angle OCX=30^{\circ}$$

