calculating angle of fitting a rectangle inside a larger one such that 3 of its corners touch the side of the larger one

Introduction

this is a useful exercise for camping when one needs to ascertain at what angle a bed or rectangular inner tent can be placed inside a larger tent

let the sides of the larger rectangle measure a and b while the lengths of the smaller rectangle measures c and d
the diagonal of the smaller one must be less than the diagonal of the larger one to be able to be rotated inside it ie sqrt(c² + d²) < sqrt(a² + b²)
to touch 3 sides, the diagonal of the smaller one must be GREATER than the length of the SHORTEST side of the larger rectangle, if it is smaller, it can be rotated more but it will only touch two sides at most
the angle of the smaller rectangles diagonal is alpha and tan(alpha) = c/d
in a coordinate setup, place the larger rectangle as [0,a] and [0,b] rotate the smaller rectangle about a chosen point, and impose that three of the rotated vertices land on the edges x=0, x=a, y=0, or y=b.
the small rectangle will be rotated by angle theta such that 3 of its corners touch the side of the larger one
horizontal projection of smaller triangle (H) = c cos theta + d sin theta
vertical projection of smaller triangle (V) = c sin theta + d cos theta
where theta is positive and < 90deg
if one wished to solve for all 4 corners touching the sides, H = a and V = b, but for 3 corners we will just use H = a hence, a = c cos theta + d sin theta
rearrange using trigonometric auxiliary angle identity:
- a = sqrt{c² + d²} x sin(theta + alpha)
solve for theta
- theta = arcsin(a/sqrt{c² + d²}) - arctan(c/d)

large tent is a = 2.69m, b = 4.42m ⇒ diagonal = 5.2m
small tent is c = 1.55m, d = 2.2m ⇒ diagonal = 2.69m, hence it can be rotated inside; the diagonal is at an angle alpha = 35.17deg
this is a special case where the diagonal of the small tent = shortest side of the large tent
- in this case theta will be 90 - alpha = 54.8deg