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

(i) Given,

⇒ 5 cos θ - 12 sin θ = 0

⇒ 5 cos θ = 12 sin θ

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

⇒ tan θ = $\dfrac{5}{12}$.

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

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