By Pythagoras Theorem:

${\displaystyle |\mathrm{AB}|=\sqrt{3},\hspace{0.5em}|\mathrm{BC}|=\sqrt{2},\hspace{0.5em}|\mathrm{AC}|=\sqrt{5},\hspace{0.5em}|\mathrm{BD}|=\sqrt{5},\hspace{0.5em}|\mathrm{AD}|=\sqrt{1}0.}$

As $|\mathrm{AB}{|}^2+|\mathrm{BC}{|}^2=|\mathrm{AC}{|}^2$ it follows, by the converse of Pythagoras Theorem, that $\angle \mathrm{ABC}={90}^o$.

Using the cosine rule: $|\mathrm{AD}{|}^2=|\mathrm{AB}{|}^2+|\mathrm{BD}{|}^2-2|\mathrm{AB}||\mathrm{BD}|\cos B.$ So

${\displaystyle \cos B=\frac{-1}{\sqrt{1}5}}$

and $\angle \mathrm{ABD}=105$ degrees to the nearest degree.

The challenge was to find more than one method so the second method uses geometry and vectors.

Triangle $A^{*}\mathrm{BC}$ is a right angled isosceles triangle and $A^{*}B$ is perpendicular to $\mathrm{BC}$.The line $B^{*}B$ is perpendicular to the plane $A^{*}\mathrm{BDC}$ so it is perpendicular to $\mathrm{BC}$. As the two line $A^{*}B$ and $B^{*}B$ are perpendicular to $\mathrm{BC}$ the whole plane ${\mathrm{AB}}^{*}{\mathrm{BA}}^{*}$ is perpendicular to $\mathrm{BC}$. Hence $\mathrm{AB}$ is perpendicular to $\mathrm{BC}$.

Using vectors :

${\displaystyle \overrightarrow{\mathrm{BA}}=(-1,-1,-1),\hspace{0.5em}\overrightarrow{\mathrm{BD}}=(2,0,-1).}$

So we find the scalar product:

${\displaystyle \overrightarrow{\mathrm{BA}}.\overrightarrow{\mathrm{BD}}=|\mathrm{BA}||\mathrm{BD}|\cos B}$

which gives

${\displaystyle -2+1=\sqrt{3}\sqrt{5}\cos B.}$

Again:

${\displaystyle \cos B=\frac{-1}{\sqrt{1}5}}$

and $\angle \mathrm{ABD}=105$ degrees to the nearest degree.