The last paragraph at the Mathworld piece on equilateral triangles gives the answer, and cites Madachy, J. S. Madachy's Mathematical Recreations. New York: Dover, pp. 115 and 129-131, 1979.
EDIT (in response to request from Taha Akbari for more detail): Let the square have horizontal and vertical sides. Consider an equilateral triangle with one vertex at the lower left corner, $A$, of the square, and one vertex at the upper left corner, $B$, of the square, and the third vertex, $Z$, inside the square. Now consider moving the triangle vertex at $B$ to the right, toward the upper right corner, $C$, of the square, while moving $Z$ so as to keep the triangle equilateral. This increases the area of the triangle, since it increases the length of the side of the triangle, since the second vertex, $X$, of the triangle is moving away from the first vertex of the triangle.
Eventually, the triangle vertex $Z$ lies on the right side of the square, and you can't move $X$ any farther right without pushing $Z$ outside the square, so you've made the triangle as large as possible. Now the question is, why are the angles $BAX$ and $ZAD$ 15 degrees (where $D$ is the lower right corner of the square)?
The triangles $BAX$ and $ZAD$ are congruent, since $BA=AD$, $AX=AZ$, and the angles at $B$ and $D$ are equal. So the angles $BAX$ and $ZAD$ are equal. But they, together with the 60 degree angle $XAZ$, add up to the 90 degree angle $BAD$. So, they measure 15 degrees.
Best Answer
I believe the following diagrams and incorporated explanation will suffice. Let me know if it is not clear. Click on image to get a larger and clearer view.