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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).

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  • $\begingroup$ Please help me! $\endgroup$ – Adele Mar 18 at 20:56
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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 $$\Lambda = \sin^{-1}\left(\sqrt{\frac{3}{4}}\right) = 60^\circ$$

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  • $\begingroup$ Thank you, very much! $\endgroup$ – Adele Mar 24 at 15:48
  • $\begingroup$ @Adele If you found this answer helpful, you can upvote it and/or accept it. $\endgroup$ – Yly Mar 24 at 17:44
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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 $DAB$.

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