I saw this question in an DSE exam.
I can solve it via:-
Method #1 (Numerical method)
(1.1) Letting PQ = 1; (1.2) Turn all lines into numerical data (eg. $QR= \dfrac {1}{ \tan 47^0} = 0.932515$); (1.3) Apply cosine law to get he required.
Method#1.1 (Testing each options)
(1) From method #1, I got RS = 1.462918; RP = … and PS = …. (2) Use the cosine formula program to test each option.
This approach is not recommended because it only requires techniques in answering mc question and has no value in edutcation. *** the following comment is deleted ***
Method #2 (applying trigometric identities)
Essentially, this is the same as method #1 except I leave the value of each line un-simpliied (eg. $QR= \dfrac {1}{ \tan 47^0}$) so that some cancellations can be done via trigonometric identities, and special angles.
All the methods require quite a bit of computations/simplifications and I don’t think the question can be solved within 1.5 minutes for an MC question.
So is the question fair? Or is there a tricky way (like considering the volume of the tetrahefron) that can give an answer quicker?
Let $\measuredangle RPS=\alpha$.
Thus, by the law of cosines we have $$PS^2+PR^2-2PS\cdot PR\cos\alpha=SQ^2+RQ^2+SQ\cdot RQ$$ or $$2PS\cdot PR\cos\alpha=PQ^2+PQ^2-SQ\cdot RQ$$ or $$\cos\alpha=\frac{PQ^2}{PS\cdot PR}-\frac{SQ\cdot RQ}{2PS\cdot PR}$$ or $$\cos\alpha=\sin53^{\circ}\sin47^{\circ}-\frac{1}{2}\cos53^{\circ}\cos47^{\circ},$$ which gives $$\alpha=\arccos\left(\sin53^{\circ}\sin47^{\circ}-\frac{1}{2}\cos53^{\circ}\cos47^{\circ}\right)\approx68^{\circ}.$$