Given : cos A = 0.6; find all other trigonometrical ratios for angle A.

Given:

cos A = 0.6

cos $A = \dfrac{6}{10}$

cos $A = \dfrac{3}{5}$

i.e. $\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{3}{5}$

∴ If length of AB = 3x unit, length of AC = 5x unit.

In Δ ABC,

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

⇒ (5x)² = (3x)² + BC²

⇒ 25x² = 9x² + BC²

⇒ BC² = 25x² - 9x²

⇒ BC² = 16x²

⇒ BC = $\sqrt{16\text{x}^2}$

⇒ BC = 4x

sin $A = \dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{BC}{AC} = \dfrac{4x}{5x} = \dfrac{4}{5}$

tan $A = \dfrac{Perpendicular}{Base}$

$= \dfrac{BC}{AB} = \dfrac{4x}{3x} = \dfrac{4}{3} = 1\dfrac{1}{3}$

cot $A = \dfrac{Base}{Perpendicular}$

$= \dfrac{AB}{BC} = \dfrac{3x}{4x} = \dfrac{3}{4}$

cosec $A = \dfrac{Hypotenuse}{Perpendicular}$

$= \dfrac{AC}{BC} = \dfrac{5x}{4x} = \dfrac{5}{4} = 1\dfrac{1}{4}$

sec $A = \dfrac{Hypotenuse}{Base}$

$= \dfrac{AC}{AB} = \dfrac{5x}{3x} = \dfrac{5}{3} = 1\dfrac{2}{3}$

Hence, sin $A = \dfrac{4}{5}$, tan $A = 1\dfrac{1}{3}$, cot $A = \dfrac{3}{4}$, cosec $A = 1\dfrac{1}{4}$ and sec $A = 1\dfrac{2}{3}$.