图中显示了一个直角三角形 OQR，其中 OQ = 4 cm，P 是 OR 上的一点，使得 OP = 3 cm 且 PR = 6 cm。 用最简分数表示，

1. Find the hypotenuse of right-angled triangle OPQ and apply the cosine ratio for angle QPR.

In right-angled triangle $OPQ$,
$PQ=\sqrt{OP^2+OQ^2}=\sqrt{3^2+4^2}=5.$
For $\angle QPR$,
$\cos\angle QPR=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{OP}{PQ}=\frac{3}{5}.$

1. Identify the opposite and adjacent sides for angle PRQ and apply the tangent ratio.

Since $P$ lies on $OR$, $\angle PRQ = \angle ORQ$.

In right-angled triangle $OQR$,
$OR = OP + PR = 3 + 6 = 9\text{ cm}.$
For $\angle PRQ$ at $R$,
$\tan \angle PRQ = \frac{\text{opposite}}{\text{adjacent}} = \frac{OQ}{OR} = \frac{4}{9}.$

1. Find the needed hypotenuse lengths in the two right-angled triangles.

In $\triangle OPQ$,
$PQ=\sqrt{3^2+4^2}=5.$
Since $OR=OP+PR=3+6=9$, in $\triangle OQR$,
$QR=\sqrt{9^2+4^2}=\sqrt{97}.$

2. Write both sine ratios and simplify their quotient.

For $\angle PQR$ in $\triangle OPQ$,
$\sin\angle PQR=\frac{OP}{PQ}=\frac{3}{5}.$
For $\angle PRQ$ in $\triangle OQR$,
$\sin\angle PRQ=\frac{OQ}{QR}=\frac{4}{\sqrt{97}}.$
Hence
$\frac{\sin\angle PQR}{\sin\angle PRQ}=\frac{\frac{3}{5}}{\frac{4}{\sqrt{97}}}=\frac{3\sqrt{97}}{20}.$