In $\triangle PQR$, right-angled at $Q$, $PR + QR = 25 \, cm$ and $PQ = 5 \, cm$. Determine the values of $\sin P, \cos P$ and $\tan P$.

(N/A) Given that, $PR + QR = 25 \, cm$ and $PQ = 5 \, cm$.
Let $PR = x \, cm$.
Then, $QR = (25 - x) \, cm$.
Applying the Pythagoras theorem in $\triangle PQR$, we have:
$PR^2 = PQ^2 + QR^2$
$x^2 = 5^2 + (25 - x)^2$
$x^2 = 25 + 625 - 50x + x^2$
$50x = 650$
$x = 13$
So, $PR = 13 \, cm$ and $QR = 25 - 13 = 12 \, cm$.
Now, we find the trigonometric ratios for $\angle P$:
$\sin P = \frac{\text{Side opposite to } \angle P}{\text{Hypotenuse}} = \frac{QR}{PR} = \frac{12}{13}$
$\cos P = \frac{\text{Side adjacent to } \angle P}{\text{Hypotenuse}} = \frac{PQ}{PR} = \frac{5}{13}$
$\tan P = \frac{\text{Side opposite to } \angle P}{\text{Side adjacent to } \angle P} = \frac{QR}{PQ} = \frac{12}{5}$