Answer:
D. 270° < φ < 360°
Step-by-step explanation:
Imagine coordinate system
I quarter is where x>0 and y>0 {right top} and it is (0°,90°)
II quarter is where x<0 and y>0 {left top} and it is (90°,180°)
III quarter is where x<0 and y<0 {left bottom} and it is (180°,270°)
IV quarter is where x>0 and y<0 {right bottom} and it is (270°,360°)
Now, we have an angle wich vertex is point (0,0) and one of its sides is X-axis and the second lay at one of the quarters.
For the trig functons of an angle created by this second side always are true:
In first quarter all functions are >0
in second one only sine
in third one: tangent and cotangent
and in fourth one: cosine
{You can check this by selecting any point on the second side of angle and put it's coordinates to formulas of these functions:
[tex]\sin \phi=\dfrac y{\sqrt{x^2+y^2}}\,,\quad \cos \phi=\dfrac x{\sqrt{x^2+y^2}}\,,\quad \tan\phi=\dfrac yx\,,\quad \cot\phi=\dfrac xy[/tex] }
So:
sinφ<0 ⇒ III or IV quarter
tanφ<0 ⇒ I or IV quarter
IV quarter ⇒ φ ∈ (270°, 360°)
