Question

Can someone please tell me how to solve this problem??!! I literally have to go back in math if I don’t pass this HELP!!

Answer 1

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:

$\sin \phi=\dfrac y{\sqrt{x^2+y^2}}\,,\quad \cos \phi=\dfrac x{\sqrt{x^2+y^2}}\,,\quad \tan\phi=\dfrac yx\,,\quad \cot\phi=\dfrac xy$  }

So:

sinφ<0  ⇒ III or IV quarter

tanφ<0  ⇒ I or IV quarter

IV quarter  ⇒  φ ∈ (270°, 360°)