Question 1015529
<pre>
In the figure below we calculate the hypotenuse by the
Pythagorean theorem using ADJACENT as the x-coordinate
of (-3,4), which is -3, and using OPPOSITE as the y-coordinate
of (-3,4), which is 4:

{{{HYPOTENUSE^2}}}{{{""=""}}}{{{ADJACENT^2+OPPOSITE^2}}}
{{{HYPOTENUSE^2}}}{{{""=""}}}{{{(-3)^2+(""+4)^2}}}
{{{HYPOTENUSE^2}}}{{{""=""}}}{{{9+16}}}
{{{HYPOTENUSE^2}}}{{{""=""}}}{{{25}}}
{{{HYPOTENUSE}}}{{{""=""}}}{{{sqrt(25)}}}
{{{HYPOTENUSE}}}{{{""=""}}}{{{5}}}



{{{drawing(600,600,-5,5,-5,5,
red(arc(0,0,3,-3,0,127)),
graph(600,600,-5,5,-5,5),locate(-2,3,HYPOTENUSE),
red(locate(.5,1.65,matrix(1,2,ANGLE,theta))),
locate(-1.6,2.7,"=+5"),locate(-3-.4,4+.4,"(-3,4)"),

locate(-2.5,.34,ADJACENT=-3),locate(-4.7,2,OPPOSITE=""+4),
green(line(0,0,-3,0),line(-3,0,-3,4),line(0,0,-3,4)) )}}}

Then we use the definition of the 6 trig ratios, being sure
to watch the signs + and -

{{{sin(theta)}}}{{{""=""}}}{{{OPPOSITE/HYPOTENUSE}}}{{{""=""}}}{{{(""+4)/(""+5)}}}{{{""=""}}}{{{4/5}}}

{{{cos(theta)}}}{{{""=""}}}{{{ADJACENT/HYPOTENUSE}}}{{{""=""}}}{{{(""-3)/(""+5)}}}{{{""=""}}}{{{-3/5}}}

{{{tan(theta)}}}{{{""=""}}}{{{OPPOSITE/ADJACENT}}}{{{""=""}}}{{{(""+4)/(""-3)}}}{{{""=""}}}{{{-4/3}}}

{{{sec(theta)}}}{{{""=""}}}{{{HYPOTENUSE/ADJACENT}}}{{{""=""}}}{{{(""+5)/(""-3)}}}{{{""=""}}}{{{-5/3}}}

{{{csc(theta)}}}{{{""=""}}}{{{HYPOTENUSE/OPPOSITE}}}{{{""=""}}}{{{(""+5)/(""+4)}}}{{{""=""}}}{{{5/4}}}

{{{cot(theta)}}}{{{""=""}}}{{{ADJACENT/OPPOSITE}}}{{{""=""}}}{{{(""-3)/(""+4)}}}{{{""=""}}}{{{-3/4}}}

Edwin</pre>