In △PQR, ∠Q = 90°. If PQ = 40 cm and PR + QR = 50 cm, find :

(i) sin P

(ii) cos P

(iii) tan R

In right ∆PQR,

∠Q = 90°

PQ = 40 cm

Given,

⇒ PR + QR = 50 cm

⇒ PR = 50 - QR

Using Pythagoras theorem we get,

⇒ PR² = PQ² + QR²

⇒ (50 - QR)² = (40)² + QR²

⇒ 50² + QR² - 100QR = 1600 + QR²

⇒ QR² - QR² - 100QR = 1600 - 2500

⇒ -100QR = -900

⇒ 100QR = 900

⇒ QR = $\dfrac{900}{100}$ = 9.

∴ PR = 50 - QR = 50 - 9 = 41.

(i) sin P = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\dfrac{QR}{PR} = \dfrac{9}{41}$.

Hence, sin P = $\dfrac{9}{41}$.

(ii) cos P = $\dfrac{\text{Base}}{\text{Hypotenuse}}$

= $\dfrac{PQ}{PR} = \dfrac{40}{41}$.

Hence, cos P = $\dfrac{40}{41}$.

(iii) tan R = $\dfrac{\text{Perpendicular}}{\text{Base}}$

= $\dfrac{PQ}{QR} = \dfrac{40}{9}$.

Hence, tan R = $\dfrac{40}{9}$.