From the figure (2) given below, find the values of :

(i) 5 sin x

(ii) 7 tan x

(iii) 5 cos x - 17 sin y - tan x

(i) By formula,

$\text{sin x } = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] = \dfrac{AD}{AB} = \dfrac{15}{25} = \dfrac{3}{5}. \\[1em] 5\text{ sin x} = 5 \times \dfrac{3}{5} = 3.$

Hence, 5 sin x = 3.

(ii) In right angle triangle ABD,

⇒ AB² = AD² + BD²

⇒ 25² = 15² + BD²

⇒ 625 = 225 + BD²

⇒ BD² = 625 - 225

⇒ BD² = 400

⇒ BD = $\sqrt{400}$ = 20.

By formula,

$\text{tan x } = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] = \dfrac{AD}{BD} = \dfrac{15}{20} = \dfrac{3}{4}. \\[1em] 7\text{ tan x} = 7 \times \dfrac{3}{4} = \dfrac{21}{4} = 5\dfrac{1}{4}.$

Hence, 7 tan x = $5\dfrac{1}{4}$.

(iii) In right angle triangle ADC,

⇒ AC² = AD² + DC²

⇒ 17² = 15² + DC²

⇒ 289 = 225 + DC²

⇒ DC² = 289 - 225

⇒ DC² = 64

⇒ DC = $\sqrt{64}$ = 8.

Solving,

$\text{5 cos x - 17 sin y - tan x} = 5 \times \dfrac{BD}{AB} - 17 \times \dfrac{CD}{AC} - \dfrac{AD}{BD} \\[1em] = 5 \times \dfrac{20}{25} - 17 \times \dfrac{8}{17} - \dfrac{15}{20} \\[1em] = 4 - 8 - \dfrac{3}{4} \\[1em] = -4 - \dfrac{3}{4} \\[1em] = \dfrac{-16 - 3}{4}\\[1em] = -\dfrac{19}{4} \\[1em] = -4\dfrac{3}{4}.$

Hence, 5 cos x - 17 sin y - tan x = -$4\dfrac{3}{4}.$