Two Triangles

I have a plastic triangle with a 45-degree angle. I lay a square, flat piece of cardboard on it and now the cardboard is at a 45-degree angle with the table. I have a second plastic triangle with a 30-degree angle. I place this triangle under the cardboard, such that it rests on the table, next to the 45-degree triangle. Now, the top edge of the 30-degree triangle doesn't reach the cardboard, so I rotate it around the 30-degree vertex until the top edge is against the cardboard, the bottom edge is on the table, and the vertex touches the vertex of the 45-degree triangle. (At all times, the triangles are held vertical.)

The end result of this is that both triangles appear to have 45-degree angles when viewed along the edge of the cardboard.

  1. What is the angle between the two triangles (i.e. the angle on the table between the sides resting on the table)?
  2. If I were to draw a line on the cardboard along the 30-degree triangle, then raise the cardboard, what angle would I measure between the bottom of the cardboard and that line?
Can you generalize this? Given an angle A and an angle B < A, what would be the horizontal angle between them, such that they would both look like angle A when viewed perpendicularly to the plane of A?

Source: Original.

Solutions were received from several people. Here are their solutions. I haven't yet had a chance to critique them, so I leave that to you.
From Joseph W. DeVincentis: 
Assume that the unspecified parts of the triangles are such that they have
right angles at the other corners that are on the table, and let the
30-degree be long enough that (when viewed in cross-section) it has the
same size as the 45-degree triangle.

Now, let the 45-degree triangle have legs of length 1.  The vertical leg of
the 30-degree triangle is also 1.  Since it's a 30-degree right triangle,
we know its hypotenuse is 2 and the other leg is sqrt(3).  Now the triangle
on the table is a right triangle with one leg of 1 and hypotenuse sqrt(3).
The angle we want is the one between those sides, or arccos(1/sqrt(3)).

Using the same technique, replacing the 45-degree angle with A and the
30-degree angle with B, and keeping the vertical height as 1, the legs on
the table are equal to the cotangents of the vertex angles, so the angle
we want is arccos(cot(A)/cot(B)) or arccos(tan(B)/tan(A)).
From Alex Doskey: 
Another way of looking at this was to imagine that our two triangles had the
same height and set this to one unit (the 30 degree triangle would obviously
be longer).  We could then put these two height sides together to make one
edge of a tetrahedron.  One face would be the 45 degree triangle, one would
be the 30 degree triangle, one would be the cardboard between the triangles
and the last face would be the table between the triangles.  Make sure you
notice that all four faces contain a right angle.

Now let's start taking measurements:
45 triangle (angle A):
height_A = 1 (by our assumption)
length_A = 1 (same as height) = 1/tan(A)
hypotenuse_A = 1.414... = sqrt(2) = 1/sin(A)

30 triangle (angle B):
height_B = 1 (by our assumption)
length_B = 1.732... = sqrt(3) = 1/tan(B)
hypotenuse_B = 2 = 1/sin(B)

Table T:
height_T = 1 unit (same edge as length_A)
hypotenuse_T = 1.732... = sqrt(3) = 1/tan(B) (same edge as length_B)
angle T = 35.264...degrees - inverse_sin(height_T/hypotenuse_T)
length_T = 1.414... = sqrt(2) = 1/tan(T)

Cardboard C:
height_C = 1.414... = sqrt(2) (same as hypotenuse_A)
length_C = 1.414... = sqrt(2) (same as length_T)
hypotenuse_T = 2 (same as hypotenuse_B)
angle_C of course is a 45 degree angle

I solved for the specific case mentioned, but I left the math there which
should work on any angles B < A< 90.
From Robert T McQuaid: 
  For the first problem, imagine a right triangle in the
  plane of the table, with two sides at the intersection
  of the vertical plastic triangles with the table, and
  the third side at right angles to the 45 degree
  plastic.  Since it is parallel to the horizontal line
  where the cardboard touches the table, the height of
  the cardboard off the third side is constant.  Calling
  the length of the side under the 45 degree plastic d,
  the height of the cardboard off the third side is also
  d.  Now draw a vertical line from the intersection of
  the third side with the the 30 degree plastic, in the
  plane of the plastic, up to the cardboard.  That
  vertical line and the 30 degree angle form a vertical
  triangle, and it is easy to calculate the bottom length
  as d/tan(30 degrees) = d * sqrt(3).  (Its hypotenuse,
  for problem 2, is d/sin(30 degrees)).  Back to the
  horizontal triangle, we know one side is d and the
  hypotenuse is d*sqrt(3), so cos(z)=1/sqrt(3) or
  z=arccos(1/sqrt(3)), the required answer.

  It is pretty complicated to describe the next result
  without a diagram, so I will just give my general
  result from similar reasoning:
  z=arccos(tan(B)/tan(A)).

  For the second problem, using the nomenclature defined
  above, the problem amounts to solving a right triangle
  in the plane of the cardboard with the intersections of
  the two pieces of plastic and an third line in the same
  vertical plane as the third line defined above.  The
  hypotenuse of this triangle is d/sin(30 degrees) or
  d*2.  The angle requested is the complement of the
  angle at the edge of the cardboard, so sin(w)=sqrt(2)/2
  and w=arcsin(sqrt(2)/2)=45 degrees.

  With a similar cop-out about complexity, the general
  result is w=arcsin(sin(B)/(cos(A)*tan(A)).
From Arturo Pascalin: 
Angle between two triangles 60 degrees.

For the general case if we call C the horizontal angle 
between the two triangle, then:

C = 90 - arctan (tan B/tan A)
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