In the figure (1) given below, use trigonometry, find $\dfrac{DE}{AD}$ if $\dfrac{BC}{AB} = \dfrac{5}{12}$.

By formula, $\text{tan θ} = \dfrac{\text{Perpendicular}}{\text{Base}}$

In ΔABC,

$\Rightarrow \text{tan θ} = \dfrac{BC}{AB} \\[1em] \Rightarrow \text{tan θ} = \dfrac{5}{12}…………….(1)$

In ΔADE,

$\Rightarrow \text{tan θ} = \dfrac{DE}{AD}$

Substituting the value of tan θ from equation (1), we get

$\Rightarrow \dfrac{DE}{AD} = \dfrac{5}{12}$.

Hence, $\dfrac{DE}{AD} = \dfrac{5}{12}$.