Question

which of the following statements shows how to calculate the reference angle for \\( \theta = \frac{11\pi}{6} \\)? \\( \circ \pi - \theta \\) \\( \circ \theta - \pi \\) \\( \circ 2\pi - \theta \\) done

Explanation:

Step1: Determine quadrant of θ

$\theta = \frac{11\pi}{6}$. Since $\frac{3\pi}{2} < \frac{11\pi}{6} < 2\pi$, θ is in Quadrant IV.

Step2: Reference angle formula for Quadrant IV

Reference angle = $2\pi - \theta$.

Answer:

$2\pi-\theta$