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

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To determine the necessary width of a wrench to tighten a regular hexagon nut with each side measuring 9.53 mm, the law of cosines can be applied. The width between two parallel sides of the hexagon is calculated as the height of an equilateral triangle formed by the sides. Using the formula d = 2x sin(60°), where x is the side length, the width is found to be approximately 16.51 mm. Therefore, a wrench should be opened to about 16.51 mm to fit the nut properly. This calculation provides the precise measurement needed for the task.
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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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