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If $ABCD$ is a regular tetrahedron with $AB=l, $what is the measure of the angle between $AB$ and $ACD$ or mathematically $\angle(AB, (ACD))$ and I ask you, please, to explain the method. Thanks.

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  • $\begingroup$ Construct the lines $AE$ and $BE$ where $E$ is the midpoint of $CD$. $ABE$ is a triangle (isosceles) and the angle at $A$ is the one you want to find. $AE$ and $BE$ are perpendicular to $CD$ so you have plenty of information to find the angle you want. $\endgroup$ Commented Sep 2, 2012 at 20:06

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If the side length is 1, then the height of a face is $\sqrt3\over2$ by Pythagoras. Let $P$ be the projection of $B$ onto $ACD$. By symmetry it is the center of $ACD$, hence at a distance of $\frac23\cdot {\sqrt3\over2}={\sqrt3\over3}$ from $A$. Then ${\sqrt3\over3}$ is also the cosine of $\angle PAB$. Thus $\angle PAB = \arccos({\sqrt3\over3}) \approx 0.9553166181245 \approx 54.7356103172^o$.

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If you know that the cosine of the dihedral angle is $1/3$ ---and everyone should know that, because it fits a neat pattern that I describe in this answer--- then take $M$ to be the midpoint of $CD$ and consider the Law of Sines in $\triangle ABM$.

$$\frac{AB}{\sin\angle AMB} = \frac{AM}{\sin\angle ABM}$$ $$\sin\angle ABM = \frac{AM}{AB} \sin\angle AMB = \frac{\sqrt{3}}{2} \sqrt{1-\left(\frac{1}{3}\right)^2} = \frac{\sqrt{3}}{2}\frac{\sqrt{8}}{3} = \frac{\sqrt{6}}{3}$$

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Position the three points $ADC$ at the vertices of an equilateral triangle in the $xy$-plane, say $A(1,0,0)$, $D(-1/2,\sqrt{3}/2, 0)$ and $C(-1/2,-\sqrt{3}/2,0)$; then the side lengths are $\sqrt{3}$; $B$ is on the $z$-axis at an equal distance from A(1,0,0), so it is $B(0,0,\alpha)$ where $1+\alpha^2=3$.

Thus we have a right triangle $BOA$ with one leg of length 1, other of length $\sqrt{2}$ so the hypotenuse is $\sqrt{3}$ and then the angle $\theta$ at $A$ has $\cos(\theta)=1/\sqrt{3}$, $\sin(\theta)=\sqrt{2/3}$.

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The orthogonal projection of the three edges $\overline{AB},\overline{CB},\overline{DB}$ onto $\triangle ACD$ have length $\mathcal{l}/\sqrt3$.

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This gives $\cos(\alpha)=\frac1{\sqrt3}$

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Suppose we know that the volume of a regular polygon with side length $1$ is $V = 1/(6\sqrt{2})$. If the angle of interest $\angle(AB, (ACD)) = \alpha$, then $$ \cos(\frac{\pi}{2} - \alpha) = \frac{\overrightarrow{AB}}{|\overrightarrow{AB}|} \:\:\cdot \frac{\overrightarrow{CA}\times \overrightarrow{AD}}{|\overrightarrow{CA}\times \overrightarrow{AD}|} \quad = \frac{6V}{\sin \frac{\pi}{3}} $$ which the right side computes the cosine of the angle between $\overrightarrow{AB}$ and the normal vector to the face $\triangle CAD$, while the triple product in the numerator computes the volume of the parallelepiped, and the bottom is the sine of an angle of the face. Simplify we have: $$ \sin\alpha = \frac{\sqrt{6}}{3} $$ and $\alpha = \arcsin (\dfrac{\sqrt{6}}{3})$ coinciding with Hagen's answer.

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Notice, in equilateral $\triangle ACD$, the length of perpendicular say $AN$ drawn from vertex $A$ to the side $CD$ $$AN=l\cos 30^\circ=\frac{l\sqrt3}{2}$$

similarly in equilateral $\triangle BCD$, the length of perpendicular say $BN$ drawn from vertex $B$ to the side $CD$ $$BN=l\cos 30^\circ=\frac{l\sqrt3}{2}$$ Now, applying cosine rule in $\triangle ABN$ $$\cos\angle BAN=\frac{AB^2+AN^2-BN^2}{2(AB)(AN)}=\frac{l^2+\left(\frac{l\sqrt 3}{2}\right)^2-\left(\frac{l\sqrt 3}{2}\right)^2}{2(l)\left(\frac{l\sqrt 3}{2}\right)}=\frac{1}{\sqrt 3}$$ $$\angle BAN=\cos^{-1}\left(\frac{l}{\sqrt 3}\right)$$ hence, the required angle $$\angle (AB, ACD)=\color{red}{\cos^{-1}\left(\frac{l}{\sqrt 3}\right)\approx54.73561032^\circ}$$

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