To crack this Tough Nut look for pairs of similar triangles and angles adding up to 90 degrees and this leads to $\mathrm{PQRS}$ being a cyclic quadrilateral with $\mathrm{SQ}$ as diameter. You need to know that opposite angles of a cyclic quadrilateral add up to 180 degrees and the angle at the centre of a circle is twice the angle at the circumference subtended by the same arc. Call the centre of the circle $O$ the use the converse of Pythagoras Theorem to find the angle $\mathrm{POR}$ . The rest follows. We'll publish the first good solution sent in to this Tough Nut.