Another question asks for the dihedral angle of a general tetrahedron in terms of its edge lengths. I chose to answer by giving a formulas in terms of face areas (because those relations deserve to be better-known). Most-relevant here is this version, for a tetrahedron $OABC$, ...
$$W^2=X^2+Y^2+Z^2-2YZ\cos\angle OA - 2ZX\cos\angle OB-2XY\cos\angle OC \tag{$\star$}$$
where $W := |\triangle ABC|$, $X := |\triangle OBC|$, $Y := |\triangle AOC|$, $Z := |\triangle ABO|$ are face areas, and $\angle OA$, $\angle OB$, $\angle OC$ are dihedral angles along respective edges $\overline{OA}$, $\overline{OB}$, $\overline{OC}$.
In a regular tetrahedron, all face areas are equal ($W=X=Y=Z$), as are all dihedral angle measures ($\angle OA = \angle OB = \angle OC$), so $$W^2 = 3 W^2 - 6W^2\cos\angle OA \qquad\to\qquad \cos\angle OA = \frac{1}{3} \tag{$\star\star$}$$
I'll leave calculating the angle measure itself to the reader. $\square$
FYI ... The $0$-dimensional point, $1$-dimensional segment, $2$-dimensional triangle and $3$-dimensional tetrahedron have higher-dimensional counterparts, each known simply by the umbrella term "$n$-dimensional simplex" (or "$n$-simplex"). The bounding elements (counterparts of "vertices", "edges", and "faces") are called "facets", and we can write, for $n\geq 2$ ...
Fun Fact. The angle between two neighboring facets of a regular $n$-simplex is $\operatorname{arccos}\frac{1}{n}$ .
So, always remember and never forget: $\cos 60^\circ$ isn't merely "half of one"; it's the reciprocal of the equilateral triangle's dimension!
One way is to do it numerically for example with wolframalpha.
use Li Li solution
$$\frac{AD^2+CD^2-AC^2}{2AD\cdot CD}=\cos(\angle ADC)$$
$$\frac{BD^2+CD^2-BC^2}{2BD\cdot CD}=\cos(\angle BDC)$$
$$\frac{AD^2+BD^2-AB^2}{2AD\cdot BD}=\cos(\angle ADB)$$
like here, it's a example of 2 second degree equations
Best Answer
Put $C$ at the origin, $B$ on the $x$-axis, and $A$ on the $y$-axis; $D$ is in the $yz$-plane, $\overline{DA}$ is perpendicular to the $xy$-plane, $\overline{DB}$ makes a $30$º degree angle with the $xy$-plane (so that $\angle DBA = 30$º), $\angle CBD = 45$º, and what’s wanted is $\angle DCA$.
$\triangle DCB$ is an isosceles right triangle; there is no harm in assuming that $|CB| = 1$, in which case $|CD| = 1$, and $|DB| = \sqrt 2$. $\angle BAD$ is a right angle, since $\overline{DA}$ is perpendicular to the $xy$-plane, and $\angle DBA = 30$º, so $\triangle DBA$ is a $30-60-90$ right triangle, and $|DA| = \frac{|DB|}{2} = \frac{\sqrt 2}{2}$. Finally, $\triangle DCA$ is a right triangle with hypotenuse $|DC| = 1$ and leg $|DA| = \frac{\sqrt 2}{2}$, so the remaining side is $\sqrt{1^2 - \left( \frac{\sqrt 2}{2} \right)^2}= \frac{\sqrt 2}{2}$, $\triangle DCA$ is an isosceles right triangle, and $\angle DCA$ is indeed $45$º.
This is a little more than I should probably say for homework, but I’ve deliberately been a bit concise, and you’re still going to have to visualize it properly to follow the calculations.