To crack this Tough Nut look for pairs of similar triangles and angles adding up to $90$ degrees and this leads to $ PQRS $ being a cyclic quadrilateral with $ 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 $ POR $ . The rest follows.