Given 5 cos A - 12 sin A = 0, find the value of $\dfrac{\text{sin A + cos A}}{\text{2 cos A - sin A}}$

Given,

⇒ 5 cos A - 12 sin A = 0

⇒ 5 cos A = 12 sin A

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

⇒ tan A = $\dfrac{5}{12}$.

We need to find the value of

$\dfrac{\text{(sin A + cos A)}}{\text{(2cos A - sin A)}} \\[1em] [\text{Dividing numerator and denominator by cos θ}] \\[1em] \Rightarrow \dfrac{\dfrac{\text{(sin A + cos A)}}{\text{cos A}}}{\dfrac{\text{(2cos A - sin A)}}{\text{cos A}}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{cos A}}}{\dfrac{\text{2cos A}}{\text{cos A}} - \dfrac{\text{sin A}}{\text{cos A}}} \\[1em] \Rightarrow \dfrac{\text{tan A + 1}}{{2 - \text{tan A}}} \space [\because \text{tan A} = \dfrac{\text{sin A}}{\text{cos A}}] \\[1em] \text{Substituting values we get} \\[1em] \Rightarrow \dfrac{\dfrac{5}{12} + 1}{2 - \dfrac{5}{12}} \\[1em] \Rightarrow \dfrac{\dfrac{5 + 12}{12}}{\dfrac{24 - 5}{12}} \\[1em] \Rightarrow \dfrac{\dfrac{17}{12}}{\dfrac{19}{12}} \\[1em] \Rightarrow \dfrac{17 \times 12}{12 \times 19} \\[1em] \Rightarrow \dfrac{17}{19}.$

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