If \(\sqrt { 3}\)sinθ - 1cosθ = 0 , 0° < θ < 90°, then find θ.

15° 60° 45° 30°

30°

\(\sqrt { 3}\)sinθ - 1cosθ = 0 \(\frac{sinθ}{cosθ}\) = \(\frac{1}{√3}\) tanθ =  \(\frac{1}{√3}\) θ = 30°