From the following figure, find the values of :

(i) cos B

(ii) tan C

(iii) sin²B + cos²B

(iv) sin B.cos C + cos B.sin C

In Δ BAC,

⇒ BC² = AB² + AC² (∵ BC is hypotenuse)

⇒ (17)² = (8)² + AC²

⇒ 289 = 64 + AC²

⇒ AC² = 289 - 64

⇒ AC² = 225

⇒ AC = $\sqrt{225}$

⇒ AC = 15

(i) cos $B = \dfrac{Base}{Hypotenuse}$

$= \dfrac{AB}{BC}\\[1em] = \dfrac{8}{17}$

Hence, cos $B = \dfrac{8}{17}$.

(ii) tan $C = \dfrac{Perpendicular}{Base}$

$= \dfrac{AB}{AC}\\[1em] = \dfrac{8}{15}\\[1em]$

Hence, tan $C = \dfrac{8}{15}$.

(iii) sin²B + cos²B

$= \Big(\dfrac{Perpendicular}{Hypotenuse}\Big)^2 + \Big(\dfrac{Base}{Hypotenuse}\Big)^2\\[1em] = \Big(\dfrac{AC}{BC}\Big)^2 + \Big(\dfrac{AB}{BC}\Big)^2\\[1em] = \Big(\dfrac{15}{17}\Big)^2 + \Big(\dfrac{8}{17}\Big)^2\\[1em] = \dfrac{225}{289} + \dfrac{64}{289}\\[1em] = \dfrac{225 + 64}{289}\\[1em] = \dfrac{289}{289}\\[1em] = 1$

Hence, sin²B + cos²B = 1.

(iv) sin B.cos C + cos B.sin C

$= \dfrac{Perpendicular}{Hypotenuse} .\dfrac{Base}{Hypotenuse} + \dfrac{Base}{Hypotenuse} . \dfrac{Perpendicular}{Hypotenuse}\\[1em] = \dfrac{AC}{BC} .\dfrac{AC}{BC} + \dfrac{AB}{BC} . \dfrac{AB}{BC}\\[1em] = \Big(\dfrac{AC}{BC}\Big)^2 + \Big(\dfrac{AB}{BC}\Big)^2\\[1em] = \Big(\dfrac{15}{17}\Big)^2 + \Big(\dfrac{8}{17}\Big)^2\\[1em] = \dfrac{225}{289} + \dfrac{64}{289}\\[1em] = \dfrac{225 + 64}{289}\\[1em] = \dfrac{289}{289}\\[1em] = 1$

Hence, sin B.cos C + cos B.sin C = 1.