Question 1099671
<pre>
Plot the point (4,-6)

{{{drawing(300,300,-10,10,-10,10,graph(300,300,-10,10,-10,10),
circle(4,-6,.1),locate(4,-6,"(4,-6)") )}}}

Next draw a line from the origin down to that point (in green).
That's the terminal side.

{{{drawing(300,300,-10,10,-10,10,graph(300,300,-10,10,-10,10),locate(4,-6,"(4,-6)"),
circle(4,-6,.1),locate(4,-6,"(4,-6)"),green(line(0,0,4,-6)) )}}}

Now indicate the positive angle t by drawing an arc starting at
the right hand side of the x-axis and swinging around counter-
clockwise to the green line (in red):

 {{{drawing(300,300,-10,10,-10,10,graph(300,300,-10,10,-10,10),locate(4,-6,"(4,-6)"),
red(arc(0,0,3,-3,0,304),locate(-1.8,2,t)),green(line(0,0,4,-6)),
circle(4,-6,.1),locate(4,-6,"(4,-6)"),green(line(0,0,4,-6)) )}}}

Now draw a horizontal line from the point (4,-6) to the x-axis 
(in blue):

 {{{drawing(300,300,-10,10,-10,10,graph(300,300,-10,10,-10,10),
green(line(0,0,4,-6)),locate(4,-6,"(4,-6)"),
red(arc(0,0,3,-3,0,304)), blue(line(4,-6,4,0)),locate(-1.8,2,t)),
circle(4,-6,.1),locate(4,-6,"(4,-6)"),green(line(0,0,4,-6)) )}}}

Notice that you now have a right triangle.  Label the horizontal 
side of that right triangle as x equal to the x-coordinate of
the point (4,-6), x=4.  Also label the vertical (blue) side of
that right triangle as y equal to the y-coordinate of
the point (4,-6), y=-6.

 {{{drawing(300,300,-10,10,-10,10,graph(300,300,-10,10,-10,10),locate(4,-6,"(4,-6)"),
green(line(0,0,4,-6)),locate(4.1,-3,y=-6),locate(2,1.2,x=4),
red(arc(0,0,3,-3,0,304)), blue(line(4,-6,4,0)),locate(-1.8,2,t)),
locate(4.1,-3,y=-6),locate(2,1.2,x=4),green(line(0,0,4,-6)),
circle(4,-6,.1),locate(4,-6,"(4,-6)"),green(line(0,0,4,-6)) )}}} 

Next we calculate the hypotenuse of the right triangle, which is
the green side and the terminal side of angle t, by using the
Pythagorean theorem.  But we label that green side r, because we
think of it as a radius swinging around forming the angle t.

{{{c^2=a^2+b^2}}} 
{{{r^2=x^2+y^2}}}
{{{r^2=4^2+(-6)^2}}}
{{{r^2=16+36}}}
{{{r^2=52}}}
{{{r=sqrt(52)}}}
{{{r=sqrt(4*13)}}}
{{{r=2sqrt(13)}}}

(We always take r as a positive number, since it was positive
before it swung around to form the angle t). We label the green
side as {{{r=2sqrt(13)}}}.  [Didn't have much room to write it :)].

 {{{drawing(300,300,-10,10,-10,10,graph(300,300,-10,10,-10,10),
locate(-1,-2,r=2sqrt(13)),locate(4,-6,"(4,-6)"),
green(line(0,0,4,-6)),locate(4.1,-3,y=-6),locate(2,1.2,x=4),locate(4,-6,"(4,-6)")
red(arc(0,0,3,-3,0,304)), blue(line(4,-6,4,0)),locate(-1.8,2,t)),
locate(4.1,-3,y=-6),locate(2,1.2,x=4),green(line(0,0,4,-6)),
locate(-1,-2,r=2sqrt(13)),
circle(4,-6,.1),locate(4,-6,"(4,-6)"),green(line(0,0,4,-6)) )}}}

We use the REFERENCE angle to determine which sides are which. 
The REFERENCE angle is the angle INSIDE the triangle that has
one side as the x-axis.  We see that the adjacent side is x=4,
the opposite side is y=-6, and the hypotenuse is {{{r=2sqrt(13)}}}.

We are asked to find the cosine of the angle t.  Now we remember
that {{{COSINE=ADJACENT/HYPOTENUSE}}}. So we write:

{{{cos(t)}}}{{{""=""}}}{{{4/(2sqrt(13))}}}{{{""=""}}}{{{2/sqrt(13)}}}

But the denominator needs to be rationalized, so we multiply by {{{sqrt(13)/sqrt(13)}}}.

{{{cos(t)}}}{{{""=""}}}{{{2sqrt(13)/(sqrt(13)sqrt(143))}}}

{{{cos(t)}}}{{{""=""}}}{{{2sqrt(13)/13}}}

That's the simplest radical form.  But I just noticed that you are
asked to get a calculator and round it to 2 decimal places.  So we
get 0.5547001962 and round it to 0.55.

Edwin</pre>