Ilovethepopper's question at Yahoo Answers concerning the width of a hexagon

[MarkFL]

Here is the question:

Math problem, law of cosines.?

this is my question :
A nut is in the shape of a regular hexagon ( six sides ). If each side is 9.53 mm, what opening on a wrench is necessary to tighten the nut, use the law of cosine to solve it and thank you

Here is a link to the question:

Math problem, law of cosines.? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello ilovethepopper,

If we have a regular hexagon whose 6 sides are $x$ units in length, then to find the width of the hexagon, i.e., the distance $d$ between two parallel sides, then we merely need to double the height of an equilateral triangle whose sides are $x$.

\(\displaystyle d=2x\sin(60^{\circ})=\sqrt{3}x\)

In this problem, we are given:

\(\displaystyle x=9.53\text{ mm}\)

and so the wrench needs to be opened to a width of:

\(\displaystyle d=\sqrt{3}\cdot9.53\text{ mm}\approx16.51\text{ mm}\)