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

Given, In triangle PQR,

PQ = 5 cm , PR + QR = 25 cm

Let us say, QR = x

Then, PR = 25 – QR = 25 – x

Using Pythagoras' theorem:

PR² = PQ² + QR²

Now, substituting the value of PR, PQ, and QR, we get;

=> (25 – x)² = (5)² + (x)²

=> 25² + x² – 50x = 25 + x²

=> 625 – 50x = 25

=> 50x = 600

=> x = 12

So, QR = 12 cm

PR = 25 – QR = 25 – 12 = 13 cm

Therefore, sinP = $\frac{\mathrm{QR}}{\mathrm{PR}} = \frac{12}{13}$