In $\triangle OPQ$,right-angled at $P$,$OP = 7\, cm$ and $OQ - PQ = 1\, cm$. Determine the values of $\sin Q$ and $\cos Q$.

(N/A) In $\triangle OPQ$,by the Pythagorean theorem,we have:
$OQ^2 = OP^2 + PQ^2$
Given $OQ - PQ = 1\, cm$,so $OQ = 1 + PQ$.
Substituting this into the Pythagorean equation:
$(1 + PQ)^2 = OP^2 + PQ^2$
$1 + PQ^2 + 2PQ = OP^2 + PQ^2$
$1 + 2PQ = OP^2$
Since $OP = 7\, cm$,we have:
$1 + 2PQ = 7^2$
$1 + 2PQ = 49$
$2PQ = 48$
$PQ = 24\, cm$
Now,$OQ = 1 + PQ = 1 + 24 = 25\, cm$.
Therefore,$\sin Q = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{OP}{OQ} = \frac{7}{25}$.
And $\cos Q = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{PQ}{OQ} = \frac{24}{25}$.