Use the given figure to find :

(i) sin x°

(ii) cos y°

(iii) 3 tan x° - 2 sin y° + 4 cos y°

In Δ BCD,

⇒ BD² = BC² + CD² (∵ BD is hypotenuse)

⇒ BD² = 6² + 8²

⇒ BD² = 36 + 64

⇒ BD² = 100

⇒ BD = $\sqrt{100}$

⇒ BD = 10

In Δ ACD,

⇒ AD² = AC² + CD² (∵ AD is hypotenuse)

⇒ 17² = AC² + 8²

⇒ 289 = AC² + 64

⇒ AC² = 289 - 64

⇒ AC² = 225

⇒ AC = $\sqrt{225}$

⇒ AC = 15

(i) sin $x° = \dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{DC}{AD}\\[1em] = \dfrac{8}{17}$

Hence, sin $x° = \dfrac{8}{17}$.

(ii) cos $y° = \dfrac{Base}{Hypotenuse}$

$= \dfrac{BC}{BD}\\[1em] = \dfrac{6}{10}\\[1em] = \dfrac{3}{5}$

Hence, cos $y° = \dfrac{3}{5}$.

(iii) 3 tan x° - 2 sin y° + 4 cos y°

tan $x° = \dfrac{Perpendicular}{Base}$

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

sin $y° = \dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{CD}{BD}\\[1em] = \dfrac{8}{10}\\[1em] = \dfrac{4}{5}$

cos $y° = \dfrac{Base}{Hypotenuse}$

$= \dfrac{BC}{BD}\\[1em] = \dfrac{6}{10}\\[1em] = \dfrac{3}{5}$

Now, 3 tan x° - 2 sin y° + 4 cos y°

$= 3 \times \dfrac{8}{15} - 2 \times \dfrac{4}{5} + 4 \times \dfrac{3}{5}\\[1em] = \dfrac{24}{15} - \dfrac{8}{5} + \dfrac{12}{5}\\[1em] = \dfrac{24}{15} + \dfrac{- 8 + 12}{5}\\[1em] = \dfrac{24}{15} + \dfrac{4}{5}\\[1em] = \dfrac{24}{15} + \dfrac{4 \times 3}{5 \times 3}\\[1em] = \dfrac{24}{15} + \dfrac{12}{15}\\[1em] = \dfrac{24 + 12}{15}\\[1em] = \dfrac{36}{15}\\[1em] = \dfrac{12}{5}\\[1em] = 2\dfrac{2}{5}$

Hence, 3 tan x° - 2 sin y° + 4 cos y° = $2\dfrac{2}{5}$.