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'Strange Rectangle' printed from https://nrich.maths.org/
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.