# A problem with tetrahedron [closed]

Let ABCD be a tetrahedron with the property that opposite edges are equal. We know that the angle between the planes ABD and BCD is $$90^\circ$$ and the angle between (BCD) and (CAD) is $$60^\circ$$. Calculate the angle measure between (CAD) and (ABD).

## closed as off-topic by Abcd, Xander Henderson, Eevee Trainer, Lee David Chung Lin, Alex ProvostMar 19 at 4:30

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The hypothesis that the tetrahedron has equal opposite edges means that opposite dihedral angles are also equal, and that all faces of the tetrahedron are congruent. A consequence of this is that for any edge $$E$$, the corresponding dihedral angle $$\Lambda$$ satisfies $$\sin(\Lambda)/|E| =$$ constant, where the constant is independent of which edge we choose. (To prove this, note that the height of the tetrahedron is $$h\sin(\Lambda)$$, where $$h$$ is an altitude of a face $$F$$ and $$\Lambda$$ is the dihedral angle between that face and the base of the tetrahedron, and then note that $$h=A/|E|$$, where $$A$$ is the area of any of the faces and $$E$$ is the edge joining face $$F$$ and the base.)
Choose coordinates so that $$D$$ is the origin and $$B$$ is at $$(1,0,0)$$. Then since $$(ABD)$$ and $$(BCD)$$ have dihedral angle $$90^{\circ}$$, we can put $$A$$ somewhere in the $$xy$$ plane and $$C$$ somewhere in the $$xz$$ plane.
Since $$(DBA)$$ and $$(DBC)$$ are congruent triangles (modulo inversion), $$A$$ and $$C$$ have coordinates $$(u,v,0)$$ and $$(1-u,0,v)$$, respectively, for some $$u$$ and $$v$$. Since the length of $$AC$$ equals the length of $$DB$$, which is 1, we have $$(1-2u)^2 + 2v^2 = 1$$.
By the first remark, we have $$\sin(60^\circ) /\sqrt{(1-u)^2 + v^2} = \sin(90^\circ) / 1 = 1$$ from the edges $$DC$$ and $$DB$$, and we similarly have $$\sin(\Lambda) /\sqrt{u^2+v^2} = 1$$ from edge $$DA$$, where $$\Lambda$$ is the angle we want to know. The first equation implies $$((1-u)^2 + v^2) = \frac{3}{4}$$. Combining with the above equation $$(1-2u)^2 + 2v^2 = 1$$ we find $$1-2u^2 = \frac{1}{2}$$, so that $$u=\frac{1}{2}$$ and thus $$v=\frac{1}{\sqrt{2}}$$. Then $$\sin(\Lambda) = \sqrt{\frac{3}{4}}$$, so $$\Lambda = \sin^{-1}\left(\sqrt{\frac{3}{4}}\right) = 60^\circ$$
You can set up a coordinate system such that: $$D=(0,0,0),\quad B=(t,0,0),\quad A=(t-a,0,b),\quad C=(a,b,0),$$ where $$t$$ must be chosen such that $$AC=BD$$. Then it is not difficult to set up normal vectors to faces $$BDC$$, $$ADC$$ and $$DAB$$: $$N_{BDC}=(0,0,1),\quad N_{DAB}=(0,1,0),\quad N_{ADC}=A\times C=(-b^2,ab,b(t-a)).$$ From these one can compute the cosine of the angle between planes $$ADC$$ and $$BCD$$: equating it to $$\cos 60°$$ will give the ratio $$a^2/b^2$$. Finally, one can compute the cosine of the angle between planes $$ADC$$ and $$DAB$$.