Size of a hexagon inscribed inside a circle

I have a ridgid bolt die which takes 1.5 diameter inch round dies. I am trying to calculate the size of a hex die which would fit into the 1.5" diameter hole. I can find formulas to calculate the size of exscribed hex and the area of the inscribed hex.
For some reason my math fails and I can not figure out how to calculate this simple problem. Of course it is a regular hexagon and all six wedges are equal. I am not sure if all three sides of each wedge are equal or is the ouside edge different?
Bill D

The across-flats dimension of a regular hexagon inscribed in a known-diameter circle is calculated as Diameter of Circle x Cosine 30 degree.

It should go without saying, but the unit of measure of the Diameter and Across-Flats dimensions are the same.

All six triangles are equilateral.
The length of the edge is the radius of the circle.
The height of the triangle is half the width of the hex.
The height of the triangle sin(60º) * the radius.
1.5 * .866 = 1.3"

All your questions may be answered above. But if you want the width of the hexagon, across the flats, then here is how you calculate it. Start with one of those six equilateral triangles. It is equilateral, so all three sides are equal to the radius of the circle, which is half it's diameter.

Now, draw a line from the center of the circle/hexagon perpendicular to the side of that equilateral triangle that is a chord of the circle. That line is half the distance across the hexagon's flats. And it bisects the equilateral triangle into two right angle triangles. And the simple formula

a^2 + b^2 = c^2

applies. Assuming that c is the radius and b is half the chord (which is also half the radius), then we want to solve for a.

a^2 = c^2 - b^2

or

a = sqrt ( c^2 - b^2 )

Where b = c/2.

You have a 1.5" circle so r and c = 0.75" so:

a = sqrt ( 0.75^2 - 0.375^2 )

a = 0.6495"

But I used only one of the equilateral triangles so a is the distance from the center of the hex to a flat. It must be multiplied by 2 to get the full distance:

2a = 1.2990"