Question 1070093
Try drawing the figure.

Central angle at the circle is 120 degrees.  Look for any of six right triangles:  hypotenuse from center to a circumscribed equilateral triangle vertex; leg from center to one of the circumscribed points; another leg from this point to that same triangle vertex.  This is a special 30-60-90 right triangle.  The short leg was given, measure of 2 cm.  


Use Pythagorean Theorem triangle formula to find the value of the other leg.  TWICE this leg is the side length of the circumscribed triangle.


Can you decide how to do that much and then go the rest of the way (to find the circumscribed equilateral triangle's altitude)?
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( side length of the circumscribed triangle should be {{{4sqrt(3)}}} ).



Another triangle drawing would be useful, but if y is the altitude, side of the equilateral triangle is {{{4sqrt(3)}}}, the altitude will cut the triangle into another pair of special 30-60-90 right triangles.
{{{y^2+(2sqrt(3))^2=(4sqrt(3))^2}}}
and you can find altitude y.