From the figure (1) given below, find the values of:

(i) sin B

(ii) cos C

(iii) sin B + sin C

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

In right-angled triangle ABC,

By Pythagoras theorem, we get

⇒ BC² = AC² + AB²

⇒ AC² = BC² - AB²

⇒ AC² = (10)² - (6)²

⇒ AC² = 100 - 36

⇒ AC² = 64

⇒ AC = $\sqrt{64}$

⇒ AC = 8.

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

= $\dfrac{AC}{BC} = \dfrac{8}{10} = \dfrac{4}{5}$.

Hence, sin B = $\dfrac{4}{5}$.

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

= $\dfrac{AC}{BC} = \dfrac{8}{10} = \dfrac{4}{5}.$

Hence, cos C = $\dfrac{4}{5}$.

(iii) sin C = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\dfrac{AB}{BC} = \dfrac{6}{10} = \dfrac{3}{5}$.

Substituting values of sin B and sin C in sin B + sin C we get :

⇒ sin B + sin C = $\dfrac{4}{5} + \dfrac{3}{5} = \dfrac{7}{5}$.

Hence, sin B + sin C = $\dfrac{7}{5}$.

(iv) sin B = $\dfrac{4}{5}$, sin C = $\dfrac{3}{5}$, cos C = $\dfrac{4}{5}$.

cos B = $\dfrac{\text{Base}}{\text{Hypotenuse}}$

= $\dfrac{AB}{BC} = \dfrac{6}{10} = \dfrac{3}{5}$.

Substituting values in equation sin B cos C + sin C cos B we get :

$= \dfrac{4}{5} \times \dfrac{4}{5} + \dfrac{3}{5} \times \dfrac{3}{5} \\[1em] = \dfrac{16}{25} + \dfrac{9}{25} \\[1em] = \dfrac{25}{25} = 1.$

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