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The solution to this tough nut was submitted by Ray who does not reveal his school.
By Pythagoras' Theorem: $$AB=\sqrt 3,\ BC=\sqrt 2,\ AC=\sqrt 5,\ BD=\sqrt 5,\ AD=\sqrt{10}.$$ As $AB^2 + BC^2 = AC^2$ it follows, by the converse of Pythagoras' Theorem, that $\angle ABC = 90^o$.
Using the cosine rule: $AD^2 = AB^2 + BD^2  2ABBD\cos B.$ So $$\cos B= {1\over \sqrt{15}}$$ and $\angle ABD = 105$ degrees to the nearest degree.
The challenge was to find more than one method so the second method uses geometry and vectors.

Triangle $A^*BC$ is a right angled isosceles triangle and $A^*B$ is perpendicular to $BC$.The line $B^*B$ is perpendicular to the plane $A^*BDC$ so it is perpendicular to $BC$. As the two line $A^*B$ and $B^*B$ are perpendicular to $BC$ the whole plane $AB^*BA^*$ is perpendicular to $BC$. Hence $AB$ is perpendicular to $BC$. 
Using vectors : $$\vec{BA}=(1,1,1),\ \vec{BD} =(2,0,1).$$ So we find the scalar product: $$ \vec{BA}.\vec{BD} = BABD\cos B$$ which gives $$ 2+1 = \sqrt 3 \sqrt 5 \cos B.$$ Again: $$\cos B= {1\over \sqrt{15}}$$ and $\angle ABD = 105$ degrees to the nearest degree.