Given: 5 cos A - 12 sin A = 0; evaluate :

$\dfrac{\text{sin A} + \text{cos A }}{2\text{ cos A } - \text{sin A}}$

Given:

5 cos A - 12 sin A = 0

⇒ 5 cos A = 12 sin A

⇒ $\dfrac{\text{sin A}}{\text{cos A}} = \dfrac{5}{12}$

⇒ $\text{tan A} = \dfrac{5}{12}$

$⇒ \text{tan A} = \dfrac{Perpendicular}{Base} = \dfrac{5}{12} \\[1em]$

∴ If length of BC = 5x unit, length of AB = 12x unit.

In Δ ABC,

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

⇒ AC² = (5x)² + (12x)²

⇒ AC² = 25x² + 144x²

⇒ AC² = 169x²

⇒ AC = $\sqrt{169 \text{x}^2}$

⇒ AC = 13x

sin A = $\dfrac{Perpendicular}{Hypotenuse}$

$\dfrac{CB}{AC} = \dfrac{5x}{13x} = \dfrac{5}{13}$

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

$\dfrac{AB}{AC} = \dfrac{12x}{13x} = \dfrac{12}{13}$

Now,

$\dfrac{\text{sin A} + \text{cos A }}{2\text{ cos A } - \text{sin A}}\\[1em] = \dfrac{\dfrac{5}{13} + \dfrac{12}{13}}{2 \times \dfrac{12}{13} - \dfrac{5}{13}}\\[1em] = \dfrac{\dfrac{5 + 12}{13}}{\dfrac{24}{13} - \dfrac{5}{13}}\\[1em] = \dfrac{\dfrac{17}{13}}{\dfrac{24 - 5}{13}}\\[1em] = \dfrac{\dfrac{17}{13}}{\dfrac{19}{13}}\\[1em] = \dfrac{\dfrac{17}{\cancel{13}}}{\dfrac{19}{\cancel{13}}}\\[1em] = \dfrac{17}{19}$

Hence, $\dfrac{\text{sin A} + \text{cos A }}{2\text{ cos A } - \text{sin A}} = \dfrac{17}{19}$.