Question 1180032
The circle's radius is {{{2.3}}} units long and the terminal point is ({{{-1.35}}},{{{1.86}}})
{{{0<= theta <2pi}}}

the point  ({{{-1.35}}},{{{1.86}}}) is in Q II and tan is negative

Since the position in the 3-o'clock is basically the positive x-axis, this means the angle that is terminating at ({{{1.35}}},{{{-1.86}}}) (which is in Q IV) should look like: 

sketch

{{{ drawing( 600, 600, -5, 5, -5, 5,
circle(0,0,2.3), circle(-1.35,1.86, .1),blue(arrow(-1.35,1.86)),green(arrow(1.35,-1.86)),locate(.5,-.5,theta=54.027612950715096),green(line(1.35,0,1.35,-1.86)),
locate(-1.5,.5,highlight(theta=-54.027612950715096)),
graph( 600, 600, -5, 5, -5, 5, 0)) }}} 


Notice that we can draw a right triangle by drawing a vertical straight line from the point ({{{1.35}}},{{{-1.86}}}).

 from the right triangle we have
 
{{{tan(m)=1.86/-1.35=-1.3777777777777778}}}

a) What is the slope of the terminal ray?

{{{m=-1.3777777777777778}}}

b) Then, {{{tan^-1(m)=tan^-1(-1.3777777777777778)=-54.027612950715096}}}°


c) Does the number we get in part (b) give us the correct value of {{{theta}}}? 
 {{{ No}}}

d) Therefore, since the point  ({{{-1.35}}},{{{1.86}}}) is in Q II,
we then subtract it from {{{360}}}  because the coordinates of the point ({{{1.35}}},{{{-1.86}}}) are in the 4th quadrant, we need angle in Q II

{{{theta=360-54.027612950715096}}}°
{{{theta=305.972387049284904}}}°