Given A is an acute angle and 13 sin A = 5, evaluate :

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

Let triangle ABC be a right-angled triangle at B and A is an acute angle.

Given,

⇒ 13 sin A = 5

⇒ sin A = $\dfrac{5}{13}$

⇒ $\dfrac{BC}{AC} = \dfrac{5}{13}$

Let BC = 5k and AC = 13k.

In right angled triangle ABC,

Using pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AB² = AC² - BC²

⇒ AB² = (13k)² - (5k)²

⇒ AB² = 169k² - 25k²

⇒ AB² = 144k²

⇒ AB = $\sqrt{144k^2}$

⇒ AB = 12k.

By formula,

cos A = $\dfrac{\text{Base}}{\text{Hypotenuse}}$

= $\dfrac{AB}{AC} = \dfrac{12k}{13k} = \dfrac{12}{13}$.

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

= $\dfrac{BC}{AB} = \dfrac{5k}{12k} = \dfrac{5}{12}$.

Substituting values we get,

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

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