Question 1159156
The point ({{{-6}}}, {{{-8}}})-> angle {{{theta}}} is in quadrant{{{ III}}}

draw the triangle in that quad, the hypotenuse would be 

{{{hypotenuse=sqrt((-6)^2+(-8)^2)=sqrt(36+64)=sqrt(100)=10}}}

Note in this case {{{theta = 180 + reference-angle(alpha)}}}
{{{sin(alpha)=-8/10=-4/5}}}->{{{-53.13}}}°
{{{cos(alpha)=-6/10=-3/5}}}->{{{126.9}}}°
{{{ tan(alpha)=-8/-6=4/3}}}->{{{53.13}}}°

 Note:

{{{ -sin (alpha) = sin( theta)}}}
{{{ -cos( alpha )= cos (theta) }}}
{{{ tan( alpha) = tan (theta)}}}

 that is, the sin and cos are {{{negative}}} in the third quadrant and the tan is positive

{{{ tan( alpha)=53.13}}}°, then

 {{{theta = 180 + 53.13=233.13}}}° is angle in quadrant{{{ III}}}