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

Given,

⇒ PR + QR = 25 cm

⇒ PR = (25 - QR) cm

In right angle triangle PQR,

By pythagoras theorem,

⇒ PR² = PQ² + QR²

⇒ (25 - QR)² = 5² + QR²

⇒ 625 + QR² - 50QR = 25 + QR²

⇒ 50QR = 625 - 25 + QR² - QR²

⇒ 50QR = 600

⇒ QR = $\dfrac{600}{50}$ = 12

⇒ PR = 25 - QR = 25 - 12 = 13

We know that,

sin P = $\dfrac{\text{Side opposite to ∠P}}{\text{Hypotenuse}} = \dfrac{QR}{PR} = \dfrac{12}{13}$

cos P = $\dfrac{\text{Side adjacent to ∠P}}{\text{Hypotenuse}} = \dfrac{PQ}{PR} = \dfrac{5}{13}$

tan P = $\dfrac{\text{Side opposite to ∠P}}{\text{Side adjacent to ∠P}} = \dfrac{QR}{PQ} = \dfrac{12}{5}$

Hence, sin P = $\dfrac{12}{13}, \text{ cos P} = \dfrac{5}{13} \text{ and tan P} = \dfrac{12}{5}$.