If $sin\theta = √3cos \theta $, then what is the value of $\frac{3sin^2\theta +cos\theta}{2cos\theta +5}$ ?
Answer & explanation
Correct answer: option 3
$sin\theta = √3cos \theta $
$\frac{sin\theta}{cos\theta}$ = $\sqrt{3}$
$tan\theta = \sqrt{3}$ ⇒ $\theta = 60°$
Thus, $\frac{3sin^2\theta +cos\theta}{2cos\theta +5}$
= $\frac{3sin^260° +cos60°}{2cos60° +5}$
= $\frac{3.\frac{3}{4} + \frac{1}{2}}{2.\frac{1}{2} + 5}$
= $\frac{11}{24}$