The possible shapes of the fourth face of a tetrahedron with three right-triangular faces. Can it be: 
A.An Equilateral triangle
B.An Isosceles right triangle
 C. A Obtuse triangle?
(The three known right angle doesn't have to share a vertex.)
My teacher showed me that it first two are correct by connecting various dots on a cube, while the last one is false because "it can't be drawn".
I'm not quite satisfied with the method used, as I can't be sure whether the scenario is actually impossible or I just can't come up with one. Is there a more systematic way to check the shape of the fourth face?
 A: Option C is possible too, see the diagram below.

A: Visualize this. The base triangle is an obtuse isosceles triangle ABC with AB = BC and angle ABC = 100 deg. 
A 4th vertex D is directly vertical above point A such that triangles BAD and CAD are right triangles, angle BAD = angle CAD = 90 deg. 
If point D is high enough, angle BDC will be acute. Reduce the height of D until it becomes a right angle. 
This will happen eventually as angle BDC becomes 100 deg as the height of D tends to zero. Hence a tetrahedron with 3 right triangles and an obtuse triangle. 
A: I am assuming that the right angles are all at the same vertex, say $O$, of the tetrahedron.  Placing the remaining vertices $A,B,C$ on respectively the $x$, $y$, and $z$ axes, we obtain $A=(a,0,0)$, $B=(0,b,0)$, and $C=(0,0,c)$.  Then the sign of the cosine of the angle at $A$ for example is the same as the sign of the dot product $AB\cdot AC=a^2>0$.  Thus the angle at $A$ is acute, and similarly for the angles at $B$ and $C$.
Therefore only option (A) is possible since it's the only acute triangle on the list.
