From the following figure, find the values of :

(i) sin A

(ii) cos A

(iii) cot A

(iv) sec C

(v) cosec C

(vi) tan C.

In Δ ABC,

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

⇒ AC² = 3² + 4²

⇒ AC² = 9 + 16

⇒ AC² = 25

⇒ AC = $\sqrt{25}$

⇒ AC = 5

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

$= \dfrac{CB}{AC}\\[1em] = \dfrac{4}{5}\\[1em]$

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

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

$= \dfrac{AB}{AC}\\[1em] = \dfrac{3}{5}\\[1em]$

Hence, cos $A = \dfrac{3}{5}$.

(iii) cot $A = \dfrac{Base}{Perpendicular}$

$= \dfrac{AB}{BC}\\[1em] = \dfrac{3}{4}\\[1em]$

Hence, cot $A = \dfrac{3}{4}$.

(iv) sec $C = \dfrac{Hypotenuse}{Base}$

$= \dfrac{AC}{AB}\\[1em] = \dfrac{5}{4}\\[1em] = 1\dfrac{1}{4}\\[1em]$

Hence, sec $A = \dfrac{5}{4} = 1\dfrac{1}{4}$.

(v) cosec $C = \dfrac{Hypotenuse}{Perpendicular}$

$= \dfrac{AC}{AB}\\[1em] = \dfrac{5}{3}\\[1em] = 1\dfrac{2}{3}\\[1em]$

Hence, cosec $C = \dfrac{5}{3} = 1\dfrac{2}{3}$.

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

$= \dfrac{AB}{BC}\\[1em] = \dfrac{3}{4}\\[1em]$

Hence, tan $C = \dfrac{3}{4}$.