



 
Hi Chuck, I drew a diagram from my reading of your description. All the dimensions are in inches. The points $A, B$ and $C$ lie in a base plane and the points $D, E$ and $F$ lie in a top plane parallel to the base plane. The vertical distance between these two planes is 14 1/4 inches. I redrew the diagram with a few extra points labeled. Imagine that I put a Cartesian coordinate system in the diagram with the origin at $B$, $BC$ on the Xaxis, $BA$ on the Yaxis and the Zaxis upwards. The point $G$ is on the base plane (the XY plane) directly below $B$ and $EH$ is perpendicular to the Xaxis. Thus the length of $EH$ is 21 inches and the length of $EG$ is 14.25 inches. The angle $EGH$ is a right angle and hence using Pythagoras Theorem I get the length of $GH$ to be 15.43 inches. This is the Ycoordinate of the point $E.$ In a similar fashion I found that the Xcoordinate of $E$ is 10.27 inches. Hence the vector $BE$ can be written $(10.27, 15.43, 14.25).$ The vector $BC$ can also be written $(29 7/16, 0, 0).$ The cross product of the vectors $BC$ and $BE$ gives a vector perpendicular to the plane containing $B, C, D$ and $E$ (the front panel) which can be normalized to length 1 to obtain a unit vector $n_1} perpendicular to the front panel. I used the command Cross({29 7/16, 0, 0},{10.27, 15.43, 14.25}) in Wolfram Alpha to make this calculation and then normalized the result to obtain \[n_1 = (0., 0.678458, 0.734639) \] In a similar fashion I obtained \[n_2 = (0.811264, 0., 0.58468)\] a unit vector perpendicular to the side panel. Now look at the point $B$ from below. The angle $PQR$ is the angle between $n_1$ and $n_2$ can be found again using Wolfram Alpha with the command Arccos({0.811264, 0., 0.58468}.{0., 0.678458, 0.734639})*180/Pi The result is that the measure of the angle $PQR$ is 64.5623 degrees, and hence the measure of the angle $BQR$ is $\frac{64.5623}{2}= 32.2812$ degrees and the measure of the angle $RBQ$ is $90  32.2812 = 57.7186$ degrees. Harley 



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