Question 1077232
{{{
drawing( 300, 300, -3, 1, -3, 1,
  grid( 0 ),
  circle( 0, 0, sqrt(5) ),
  line( -1, 0, 0, 0),
  line(-1,-2, 0, 0),
  line(-1,-2, -1, 0),
  locate(-0.5,0.3, a),
  locate(-1.2,-1, b),
  locate(-0.5,-1.1, h),
  locate(-0.3,-0.1,x),
  locate(-1.2,-2, "(a,b)")
)
}}}
—

{{{tan(x) = 2}}} —>  {{{ x=tan^-1(2) = 63.4349^o }}}  and since {{{ tan(x) = b/a }}}:  {{{ a=-1 }}}, {{{b=-2}}} 
But since we're in Q3, we need to add {{{180^o}}}  to get it into standard form (where the angle is measured from the +x axis counterclockwise).  

(iii)  {{{ highlight(x=243.4349^o) }}} in standard form.

{{{ h = sqrt(a^2 + b^2) = sqrt((-1)^2+(-2)^2) = sqrt(5) }}} 

(i) {{{ cos(x) = a/h = highlight(-1/sqrt(5)) }}} or approx. -0.447214
(ii) {{{ sin(x) = b/h = highlight(-2/sqrt(5)) }}} or approx. -0.894427

===
NOTE:  Just be aware:  often you will want to normalize things so you have a unit circle (i.e. r=1).  I did not use a unit circle (h=sqrt(5)).   To do that I would have had to divide a,b, and h by sqrt(5).  However, since the problem only dealt with ratios of the sides, the lack of normalization didn't matter.