I've attached a perspective sketch of the folded triangle.
1. You already know:
$\displaystyle |\overline{BM}|= |\overline{CM}|$
$\displaystyle \angle(BMC) = 90^\circ$
2. Draw the line $\displaystyle \overline{BC}$. It is the hypotenuse of the isosceles right triangle BMC.
3. Construct the midpoint of BC = M'
4. The length of $\displaystyle \overline{MM'}$ ist calculated by
$\displaystyle |\overline{MM'}| = \frac12 \sqrt{2} \cdot |\overline{BM}|$
5. Since $\displaystyle |\overline{AB'}| = 2 \cdot |\overline{B'M}|$ the line segment $\displaystyle \overline{AM}$ has the length
$\displaystyle |\overline{AM}| = \sqrt{3} \cdot |\overline{BM}|$
6. The angle $\displaystyle \alpha$ is the angle between the plane BMC and the plane ABC. It can be calculated by
$\displaystyle \tan(\alpha) = \frac{|\overline{AM}|}{|\overline{MM'}|} = \frac{\sqrt{3} \cdot |\overline{BM}|}{\frac12 \sqrt{2} \cdot |\overline{BM}|} = \sqrt{6}$
And therefore the angle $\displaystyle \alpha \approx 67.79^\circ$