If 13 sin θ = 5, find the value of

$\dfrac{\text{5 sin θ - 2 cos θ}}{\text{tan θ}}$

13 sin θ = 5

sin θ = $\dfrac{5}{13}$

sin θ = $\dfrac{\text{perpendicular}}{\text{hypotenuse}} =\dfrac{\text{5}}{\text{13}}$

Let perpendicular = 5x and hypotenuse = 13x

By pythagoras theorem, we get :

Hypotenuse² = Base² + Perpendicular²

Base² = Hypotenuse² - Perpendicular²

Base² = (13x)² - (5x)²

Base² = 169x² - 25x²

Base² = 144x²

Base = $\sqrt{144x^2}$

Base = 12x

cos θ = $\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{12x}{13x} = \dfrac{12}{13}$

tan θ = $\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{5x}{12x} = \dfrac{5}{12}$

Substituting values, we get :

$\Rightarrow \dfrac{\text{5 sin θ - 2 cos θ}}{\text{tan θ}} = \dfrac{5 \times \dfrac{5}{13} - 2 \times \dfrac{12}{13}}{\dfrac{5}{12}} \\[1em] = \dfrac{\dfrac{25}{13} - \dfrac{24}{13}}{\dfrac{5}{12}} \\[1em] = \dfrac{\dfrac{1}{13}}{\dfrac{5}{12}} \\[1em] = \dfrac{12}{65}.$

Hence, $\dfrac{\text{5 sin θ - 2 cos θ}}{\text{tan θ}} = \dfrac{12}{65}$.