Easy 3D geometry from SAMO 2016 
The required angle can be found out by using cosine rule and then we get it as be $90^{\circ}$.
Can someone provide a bit more geometric approach? Or if trigonometric, then, not at least with any of these rules. Just normal $\sin, \cos, \tan$ would be better.
 A: This problem does not require any formulas, calculations or even exact lengths for $AB$ and $CD$. The condition $AB = CD = \sqrt{3}$ is absolutely irrelevant information.
1) Since $AB$ is orthogonal to the plane determined by the three points $B, C, D$, it is in fact orthogonal to any line in that plane. In particular, $AB$ is orthogonal to line $CD$. 
2) Furthermore, $CD$ is orthogonal to $BD$ by assumption. 
3) Since $CD$ is orthogonal to two non-parallel lines, $AB$ and $BD$, lying in the plane determined by the three points $A, B, D,$ one concludes that $CD$ is in fact orthogonal to the whole plane through $A, B, D$.
4) Since $CD$ is orthogonal to the plane determined by the three points $A, B, D$, it is in fact orthogonal to any line in that plane. In particular, $CD$ is orthogonal to line $AD$.   
5) Consequently, $\angle \, ADC = 90^{\circ}$.
A: Use the Pythagorean theorem to see that $AD = BC = 2$ and then argue that triangles $ADC$ and $ABC$ are congruent. That means that the angle $ADC$ is equal to angle $ABC$.
A: By the Pythagorean Theorem in $\Delta BDC$,
$$BC^2 = BD^2+DC^2 = 1^2 + \left(\sqrt{3}\right)^2=4$$
By the Pythagorean Theorem in $\Delta ABC$,
$$AC^2 = AB^2 + BC^2 = \left(\sqrt{3}\right)^2 + 4 = 7$$
By the Pythagorean Theorem in $\Delta ABD$, 
$$AD^2 = AB^2 + BD^2 = \left(\sqrt{3}\right)^2 + 1^2 = 4$$
In $\Delta ADC$,
$$AD^2 + DC^2 = 4 +  \left(\sqrt{3}\right)^2 = 7 = AC^2$$
hence, by the converse of the Pythagorean Theorem, $\angle ADC = 90^{\circ}$.
