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From the given dimensions one then finds that the

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Unformatted text preview: posite point is given by a common formula as 0.866a. Now, we draw a triangle containing the overlying atom and the atom at the top of the above triangle: From the properties of the various triangles, one can find the indicated lengths. From the given dimensions, one then finds that the angle between the red line and the bottom edge at the right has an included angle of 19.47°. Also, from equivalent triangles, the vertical left edge has the same length as the bottom, so that the height of the triangle is 0.913a. Equivalent triangles also give the length between the green dot and the corner of the two red lines- - the nearest neighbor distance is then 0.6123a. The upper right line is an edge of the blue triangle above, so has length a. If we now use the law of cosines 12 − 2(0.6123)2 = 2 ⋅ 1 ⋅ 0.6123 cos(ϑ ) we find that the angle is 70.5°, but since the included angle between the two...
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