By Pythagoras theorem

{AB}^{2} = 3, {BC}^{2} = 2, {AC}^{2} = 5, {BD}^{2} = 5, {AD}^{2} = 10

Method 1
In

▵ ABC

we have

{AB}^{2} + {BC}^{2} = {AC}^{2}

so, by the converse of Pythagoras Theorem,

∠ ABC = 90^{o}

Clearly $▵ ABD$ is not a right angles triangle but we can use the Cosine Rule to find $∠ ABD$ .

$\cos ∠ ABD = \frac{3 + 5 - 10}{2 \sqrt{3} \sqrt{5}} = \frac{- 1}{\sqrt{1} 5}$

so $∠ ABD = 104 . 963^{o}$ or approx. 105 degrees.

Method 2

$\vec{BA} = (- 1, - 1, - 1) \vec{BC} = (1, 0, - 1), \vec{BD} = (2, 0, - 1)$

so, using the scalar product, $\vec{BA} . \vec{BC} = 0$ and so $∠ ABC = 90^{o}$ .

For angle $ABD$ :

$\vec{BA} . \vec{BD} = - 1 = \sqrt{3} \sqrt{5} \cos ∠ ABD$

Hence

$∠ ABD = \cos^{- 1} \frac{- 1}{\sqrt{1} 5} = 105^{o}$

to the nearest degree.