Question 349029
<pre><b>
Let's draw a picture and see:

find the trig functions of angle {{{theta}}} if the terminal side of {{{theta}}} passes through the point (-12, 5).

Notice I changed the name of the angle from x to {{{theta}}} because
x stands for the first coordinate of a point.  If you use x to refer
to the angle too, you will have a conflict of variables.  That's a
no-no.

{{{drawing(500,500,-15,15,-15,15, graph(500,500,-15,15,-15,15),
locate(-14,7,"(-12,5)"), line(-12,4.9,-12,5.2), line(-12-.1,5,-12+.1,5),
green(line(0,0,-24,10)), red(arc(0,0,6,-6,0,157.3801351), locate(.5,4,theta)) )}}}

Now we draw a perpendicular line down to the x-axis:

{{{drawing(500,500,-15,15,-15,15, graph(500,500,-15,15,-15,15),
locate(-14,7,"(-12,5)"), line(-12,4.95,-12,5.1), line(-12-.1,5,-12+.1,5),
green(line(0,0,-24,10)), red(line(-12,5,-12,0)), red(arc(0,0,6,-6,0,157.3801351), locate(.5,4,theta)),locate(-9,1,x=-12),
locate(-12,2.5,y=5)


 )}}}

Next we cut off the green line beyond the right triangle, and label
it r.  We use the Pythagorean theorem to calculate r:

{{{r^2=x^2+y^2}}}
{{{r^2=(-12)^2+(5)^2}}}
{{{r^2=144+25}}}
{{{r^2=169}}}
{{{r=sqrt(169)}}}
{{{r=13}}}

{{{drawing(500,500,-15,15,-15,15, graph(500,500,-15,15,-15,15),
locate(-14,7,"(-12,5)"), line(-12,4.95,-12,5.1), line(-12-.1,5,-12+.1,5),
green(line(0,0,-12,5)), red(line(-12,5,-12,0)), red(arc(0,0,6,-6,0,157.3801351), locate(.5,4,theta)),locate(-9,1.3,x=-12),
locate(-11.5,2.5,y=5), locate(-9,5,r=13)


 )}}}

{{{matrix(3,6,

sin(theta)=y/r=5/13, "", "", "", "",csc(theta)=r/y=13/5, 
cos(theta)=x/r=(-12)/13=-12/13,  "", "", "", "",sec(theta)=r/x=13/(-12)=-13/12,
tan(theta)=y/x=5/(-12)=-5/12, "", "", "", "",cot(theta)=x/y=(-12)/5=-12/5)}}} 

Edwin</pre>