Have You Considered All Possible Configurations?

Let $D$ be the foot of the perpendicular from $D$ to $BC$ .

Case 1: $AB = AC$
Then, we must have $D$ lying within $BC$ , and $AD = 1, BC = 2, DC = 1$ . This gives us $ADC, ADB$ are isosceles right triangles, and thus $\angle BAC = 90 ^ \circ$ .

Case 2: $AC = CB$ (and equivalently, $AB = BC$ )
Case 2a: $D$ lies on the line segment $BC$
Then, $AC = 2, AD = 1$ so $\angle ACB = 30 ^ \circ$ and thus $\angle BAC = \frac{ (180 - 30) } { 2} = 75 ^ \circ$ .

Case 2b: $D$ does not lie on the segment $BC$ .
Then, $\angle DCA = 30 ^ \circ$ and so $\angle BCA = 150 ^ \circ$ and thus $\angle BAC = \frac{ (180-150) } {2} = 15 ^ \circ$ .