Answer:
The angle 2pi/3 terminate in quadrant II
Step-by-step explanation:
Quadrant I ⇒ 0 < θ < (π/2)
Quadrant II ⇒ (π/2) < θ < (π)
Quadrant III ⇒ (π) < θ < (3π/2)
Quadrant IV⇒ (3π/2) < θ < (2π)
When θ = 2π/3
∴ (π/2) < θ < (π)
So, The angle 2pi/3 terminate in quadrant II
