Use the adjoining figure and write the values of :

(i) sin x°

(ii) cos y°

(iii) 3 tan x° - 2 sin y° + 4 cos y°

In right angled triangle DBC,

Perpendicular = BC = 8 cm

Base = DB = 6 cm

Then we will find hypotenuse (CD) by pythagoras theorem,

Hypotenuse² = Base² + Perpendicular²

Hypotenuse² = 6² + 8²

Hypotenuse² = 36 + 64

Hypotenuse² = 100

Hypotenuse = 10 cm

In right angled triangle ABC,

Perpendicular = CB = 8 cm

Hypotenuse = AC = 17 cm

Let AD = m

Base (AB) = AD + DB = m + DB

By pythagoras theorem,

Base² = Hypotenuse² - Perpendicular²

(m + 6)² = 17² - 8²

m² + 36 + 12m = 289 - 64

m² + 36 + 12m = 225

m² + 12m + 36 - 225 = 0

m² + 12m - 189 = 0

m² + 21m - 9m - 189 =0

m(m + 21) - 9(m + 21) = 0

(m + 21)(m - 9) = 0

m = -21 or m = 9

Sicne, length can't be negative.

so, m = 9 cm

AB = m + 6 = 9 + 6 = 15 cm

(i) sin x° = $\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{BC}{AC} = \dfrac{8}{17}$.

(ii) cos y° = $\dfrac{\text{base}}{\text{hypotenuse}}= \dfrac{DB}{DC} = \dfrac{6}{10} = \dfrac{3}{5}$.

(iii) 3 tan x° - 2 sin y° + 4 cos y°

tan x° = $\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{BC}{AB} = \dfrac{8}{15}$

sin y° = $\dfrac{\text{perpendicular}}{\text{hypotenuse}}= \dfrac{BC}{DC} = \dfrac{8}{10} = \dfrac{4}{5}$

Putting values of tan x°, sin y°, cos y° in 3 tan x° - 2 sin y° + 4 cos y°

= $3\times\dfrac{8}{15} - 2\times\dfrac{4}{5} + 4\times\dfrac{3}{5}$

= $\dfrac{8}{5} - \dfrac{8}{5} + \dfrac{12}{5}$

= $\dfrac{12}{5} = 2\dfrac{2}{5}$.