Given : $\text{sin A} = \dfrac{3}{5}$, find :

(i) tan A

(ii) cos A

(i) Given:

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

i.e., $\dfrac{Perpendicular}{Hypotenuse} = \dfrac{3}{5}$

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

In Δ ABC,

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

⇒ (5x)² = AB² + (3x)²

⇒ 25x² = AB² + 9x²

⇒ AB² = 25x² - 9x²

⇒ AB² = 16x²

⇒ AB = $\sqrt{16\text{x}^2}$

⇒ AB = 4x

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

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

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

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

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

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