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

Let we have the triangle as:

Assuming $QR = x$

Since $PR + QR = 25 cm$, $PR = 25 - QR$

$\Rightarrow PR = 25 - x$

Using Pythagoras theorem,

$P{R}^2 = Q{R}^2 + P{Q}^2$

Substituting the 5 for PQ and (25 - x) for PR.

${\left(25- x\right)}^2= 5^2 + x^2$

${25}^2 + x^2 – 50x = 25 + x^2$

$625 + x^2- 50x -25 – x^2= 0$

$-50x = -600$

$x= \frac{-600}{-50}$

$x = 12 = QR$

Using value of QR, we can find the value of PR.

$PR = 25-QR = 25-12 = 12$

So, Now we can find the required ratios,

$Sin \left(P\right) =\frac{Opposite side}{Hypotenuse}=\frac{QR}{PR}=\frac{12}{13}$

$Cos \left(P\right) =\frac{Adjacent side}{Hypotenuse}=\frac{PQ}{PR}=\frac{5}{13}$

$\tan \left(P\right) =\frac{Opposite side}{Adjacent side}=\frac{QR}{PQ}=\frac{12}{5}$