Hexagonal Bucket

Using David's labelling, $A,B,C$ have coordinates $(5,5\sqrt{3},0)$ , $(10,0,0)$ , and $(5,-5\sqrt{3},0)$ respectively, while $D,E,F$ have coordinates $(10,10\sqrt{3},h)$ , $(20,0,h)$ and $(10,-10\sqrt{3},h)$ respectively, where $h$ is the vertical height of the basket. Planes $ABED$ and $BCFE$ have respective normal vectors $\mathbf{n}_1 = \overrightarrow{BE} \times \overrightarrow{AD} \; = \; \left(\begin{array}{c} -5h\sqrt{3} \\ -5h \\ 50\sqrt{3} \end{array}\right) \hspace{2cm} \mathbf{n}_2 = \overrightarrow{CF} \times \overrightarrow{BE} \; = \; \left(\begin{array}{c} -5h\sqrt{3} \\ 5h \\ 50\sqrt{3} \end{array}\right)$ and hence $\mathbf{n}_1 \cdot \mathbf{n}_2 \; = \; 50(h^2 + 150) \hspace{2cm} |\mathbf{n}_1|^2 = |\mathbf{n}_2|^2 \; = \; 50(2h^2 + 150)$ Thus, if $\theta$ is the acute angle between two adjacent trapezia (UK terminology), then $\cos\theta \; = \; \frac{h^2 + 150}{2h^2 + 150}$ Now $h^2 + 10^2 = |BE|^2 = 30^2 + 5^2$ , and hence $h^2 = 825$ , and hence $\cos\theta = \tfrac{13}{24}$ . Thus the dihedral angle between the two trapezia is $\pi - \theta = \boxed{2.143214899}$ .

Here is a solution with no coordinate geometry.

Toolbox:

In figure 1 we have turned the bucket upside down. From vertices $A$ and $C$ we drop perpendiculars to edge $HB$ (the blue line segments) creating two congruent right triangles. Hence, the perpendiculars meet at the same point $M$ . Then the dihedral angle in question is $\angle AMC$ .

In order to calculate the length of $AM$ , we isolate the face $ABHG$ in figure 2 and split it the way David Vreken did in his solution.

By Pythagorean theorem on right triangle $\triangle HNB$ we get $HB = \sqrt {{{30}^2} + {5^2}} = 5\sqrt {37}$ .

Next, $\triangle HNB$ and $\triangle AMB$ are similar, thus, $\frac{{AM}}{{HN}} = \frac{{AB}}{{HB}} \Rightarrow \frac{{AM}}{{30}} = \frac{{20}}{{5\sqrt {37} }} \Rightarrow AM = \frac{{120}}{{\sqrt {37} }}$

The base $ABCDEF$ of the bucket is a regular hexagon. Hence, $\angle ABC = \frac{{2\pi }}{3}$ .

Using cosine rule on $\triangle ABC$ , $A{C^2} = A{B^2} + B{C^2} - 2AB \cdot BC\cos B = 2 \cdot {20^2} - 2 \cdot {20^2} \cdot \left( { - \frac{1}{2}} \right) = 1200$ By cosine rule once again, this time on $\triangle AMC$ ( figure 1 ), $\cos \left( {\angle AMC} \right) = \frac{{A{M^2} + M{C^2} - A{C^2}}}{{2AM \cdot MC}} = \frac{{2{{\left( {\frac{{120}}{{\sqrt {37} }}} \right)}^2} - 1200}}{{2{{\left( {\frac{{120}}{{\sqrt {37} }}} \right)}^2}}} = - \frac{{13}}{{24}}$

Hence, $\angle AMC = {\cos ^{ - 1}}\left( { - \frac{{13}}{{24}}} \right) \approx \boxed{2.14}$

Split one of the isosceles trapezoids as follows:

Using the Pythagorean Theorem on the right triangle with legs $5$ and $30$ , one leg of the isosceles trapezoid is $\sqrt{5^2 + 30^2} = 5\sqrt{37}$ .

Now place the hexagonal bucket on a coordinate system so that the bottom center is at the origin, two vertices of the bottom hexagon are on the $x$ -axis, and the center of the top hexagon is on the $z$ -axis, and label the points as follows:

Since the difference of the hexagons along the $x$ -axis is $BG = 20 - 10 = 10$ , and since the leg of an isosceles trapezoid is $BE = 5\sqrt{37}$ , then by the Pythagorean Theorem on $\triangle BEG$ the height of the bucket is $EG = \sqrt{(5\sqrt{37})^2 - 10^2} = 5\sqrt{33}$ . Using the properties of equilateral triangles, the coordinates for $A$ , $B$ , $C$ , and $E$ are then $A(5, 5\sqrt{3}, 0)$ , $B(10, 0, 0)$ , $C(5, -5\sqrt{3}, 0)$ , and $E(20, 0, 5\sqrt{33})$ .

The equation of the plane with $A$ , $B$ , and $E$ is in the form of $x + by + cz = d$ . Using the coordinates for $A$ , $B$ , and $E$ gives equations $5 + 5\sqrt{3}b = d, 10 = d$ , and $20 + 5\sqrt{33}c = d$ , which solve to $b = \frac{\sqrt{3}}{3}$ , $c = -\frac{2\sqrt{33}}{33}$ , and $d = 10$ , for a planar equation of $x + \frac{\sqrt{3}}{3}y - \frac{2\sqrt{33}}{33}z = 10$ , which is equivalent to $33x + 11\sqrt{3}y - 2\sqrt{33}z = 330$ . By a similar argument, the equation of the plane through $B$ , $C$ , and $E$ is $33x - 11\sqrt{3}y - 2\sqrt{33}z = 330$ .

The acute angle between the two planar equations is $\theta = \cos^{-1} \Big(\frac{(33)(33) + (11\sqrt{3})(-11\sqrt{3}) + (-2\sqrt{33})(- 2\sqrt{33})}{\sqrt{(33)^2 + (11\sqrt{3})^2 + (-2\sqrt{33})^2}\sqrt{(33)^2 + (-11\sqrt{3})^2 + (-2\sqrt{33})^2}}\Big) = \cos^{-1}(\frac{13}{24})$ , so the obtuse angle (the angle that we want) is $\pi - \theta = \pi - \cos^{-1}(\frac{13}{24}) \approx \boxed{2.143}$ .