If sin $A = \dfrac{5}{13}$ the value of tan A is :

1. $\dfrac{5}{12}$

2. $\dfrac{12}{13}$

3. $\dfrac{12}{5}$

4. $\dfrac{13}{12}$

Given:

sin $A = \dfrac{5}{13}$

i.e., $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{5}{13}$

∴ If length of BC = 5x unit, length of AC = 13x unit.

In Δ ABC,

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

⇒ (13x)² = (5x)² + BC²

⇒ 169x² = 25x² + BC²

⇒ BC² = 169x² - 25x²

⇒ BC² = 144x²

⇒ BC = $\sqrt{144x^2}$

⇒ BC = 12x

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

= $\dfrac{BC}{AB}$ = $\dfrac{5x}{12x}$ = $\dfrac{5}{12}$

Hence, option 1 is the correct option.