In the tetrahedron $ABCD$ (a pyramid with a triangular base), the edge $DA$ is perpendicular to the base $ABC$. We also know the angles between the edges meeting at vertex $D$: $\angle ADB = 30^\circ$, $\angle BDC = 45^\circ$, $\angle CDA = 60^\circ$. Construct (using the usual steps on the plane) the angle $\angle BAC$.

For students in the second grade, also calculate the angle $\angle BAC$.
1. Cut the three edges of the pyramid paper model that meet at $B$, so the net can be laid out flat on a plane. We will construct this net from the data, without the model.

![](https://cdn.mathpix.com/cropped/2024_05_02_fc910b345090dac6713dg-1.jpg?height=585&width=599&top_left_y=263&top_left_x=752)

Since (originally) $DA$ is perpendicular to the plane $ABC$, it is perpendicular to every line in the plane, so in the triangles $DAB$ and $DAC$, there is a right angle at $A$. By choosing the common length of their legs $DA$, we can construct these two triangles knowing that $\angle ADC = 60^\circ$ and $\angle ADB = 30^\circ$. Measure the angle of $45^\circ$ at $D$ on the side $DC$, and again the already obtained edge $DB$ appears on the other side of the angle, thus we have also obtained the triangle $DCB$.

Knowing the segments $BA$ and $BC$, we construct the triangle $ABC$ with $AC$ as the base, and by measuring the angle $\angle BAC$, we get approximately $50^\circ$. In the net, the vertex $B$ appears in three places, which we distinguish as $B_1, B_2, B_3$ in the order of construction.

2. Switching to calculation, let's choose the length of $AD$ as the unit and denote the projections of $B_2$ on $DC$ and $B_3$ on $AC$ as $B_2'$ and $B_3'$, respectively. Thus, based on the angle data:

$$
AB_1 = \frac{1}{\sqrt{3}}, \quad DB_1 = DB_2 = \frac{2}{\sqrt{3}}, \quad AC = \sqrt{3}, \quad DC = 2, \quad B_2B_2' = DB_2' = \frac{\sqrt{2}}{\sqrt{3}}
$$

and applying the Pythagorean theorem to the triangle $CB_2B_2'$:

$$
CB_2^2 = \left(\frac{\sqrt{2}}{\sqrt{3}}\right)^2 + \left(2 - \frac{\sqrt{2}}{\sqrt{3}}\right)^2 = \frac{16 - 4\sqrt{6}}{3}
$$

On the other hand, applying the Pythagorean theorem to the two sub-triangles of $ACB_3$:

$$
AB_3^2 - B_3'A^2 = B_3'B_3^2 = B_3C^2 - (AC - B_3'A)^2 = B_3C^2 - AC^2 + 2AC \cdot B_3'A - B_3'A^2
$$

and from this, immediately substituting the above results:

$$
B_3'A = \frac{AB_3^2 + AC^2 - B_3C^2}{2AC} = \frac{2\sqrt{2}}{3} - \frac{1}{\sqrt{3}}
$$

Thus, from the triangle $AB_3B_3'$:

$$
\cos \angle B_3AB_3' = \cos \angle BAC = \frac{B_3'A}{AB} = \frac{2\sqrt{6}}{3} - 1 = 0.6329,
$$

so the desired angle is $50^\circ 44'$.

Remarks. 1. In determining $CB_2$ and $B_3'A$, we essentially followed the path usually used to derive the cosine rule; here we also provided the intermediate results. Indeed, we could have calculated the desired data by applying the cosine rule twice (in the triangle $CDB_2$ we used $\cos 45^\circ = 1/\sqrt{2}$).

2. Our problem can also be interpreted as follows: we cut the trihedron (open solid angle) at $D$ with a plane perpendicular to the edge $DA$, so the angle between the lines $AB$ and $AC$ also measures the dihedral angle (one of the angles of the trihedron) between the planes $DAB$ and $DAC$. Similarly, we can determine the dihedral angles at the edges $DB$ and $DC$ by cutting with planes perpendicular to these edges. Accordingly, we have carried out the construction solution of the trihedron (or spherical triangle) defined by the angles $30^\circ, 45^\circ, 60^\circ$. The calculation can also be performed based on the side-cosine rule of the spherical triangle (formula 362.121 in the School Function Table).