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

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SUMMARY

The discussion addresses a mathematical problem involving the width of a regular hexagon, specifically for a nut with each side measuring 9.53 mm. To determine the necessary width for a wrench, the law of cosines is applied, leading to the formula for the distance between two parallel sides of the hexagon: d = 2x sin(60°) = √3x. Substituting the given side length, the required width for the wrench is calculated to be approximately 16.51 mm.

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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.
 
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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 $x$.

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

In this problem, we are given:

$$x=9.53\text{ mm}$$

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

$$d=\sqrt{3}\cdot9.53\text{ mm}\approx16.51\text{ mm}$$
 

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