Answer
(a) $\frac{\pi}{12}$ lies in the First Quadrant.
(b) $-\frac{11\pi}{9}$ lies in the Second Quadrant.
Work Step by Step
If $θ$ lies in the:
- First Quadrant, then $0\ltθ\lt\frac{\pi}{2}$
- Second Quadrant, then $\frac{\pi}{2}\ltθ\lt\pi$
- Third Quadrant, then $\pi\ltθ\lt\frac{3\pi}{2}$
- Fourth Quadrant, then $\frac{3\pi}{2}\ltθ\lt2\pi$
(a) $0\lt\frac{\pi}{12}\lt\frac{\pi}{2}$ , so $\frac{\pi}{12}$ lies in the First Quadrant.
(b) $-\frac{11\pi}{9}+2\pi=\frac{7\pi}{9}$
$\frac{\pi}{2}\lt\frac{7\pi}{9}\lt\pi$ , so $-\frac{11\pi}{9}$ lies in the Second Quadrant.
