By Pythagoras theorem

AB²=3, BC²=2, AC²=5, BD²=5, AD²=10

Method 1
In \triangle ABC we have AB²+BC²=AC² so, by the converse of Pythagoras Theorem, ÐABC = 90^o

Clearly \triangle ABD is not a right angles triangle but we can use the Cosine Rule to find ÐABD.

cosÐABD = 3 + 5 - 10
2Ö3 Ö5
= -1
Ö15

so ÐABD = 104.963^o or approx. 105 degrees.

Method 2

®
BA

= (-1, -1, -1) ®
BC

= (1, 0, -1), ®
BD

= (2, 0, -1)

so, using the scalar product,
®
BA

. ®
BC

= 0

and so ÐABC = 90^o.

For angle ABD:

®
BA

. ®
BD

= -1 = Ö3 Ö5 cosÐABD

Hence

ÐABD = cos^-1 -1
Ö15
= 105^o

to the nearest degree.