Question 180226
If (-1,3) are the coordinates of a point on the terminal arm of angle B in standard position, then tan B is ?

<pre><font size = 4 color = "indigo"><b>
Plot the point (-1,3) and draw a terminal arm from that point
to the origin (0,0).

{{{drawing(400,400,-4,4,-4,4, graph(400,400,-4,4,-4,4,(sqrt(.33+x)/sqrt(.33+x))*sqrt(.6-x^2)  ), locate(.3,1.1,B),locate(-1.5,3.4,"(-1,3)"), line(0,0,-1,3) )}}}

Now draw a perpendicular to the x-axis:

{{{drawing(400,400,-4,4,-4,4, graph(400,400,-4,4,-4,4,(sqrt(.33+x)/sqrt(.33+x))*sqrt(.6-x^2)  ), locate(.3,1.1,B),locate(-1.5,3.4,"(-1,3)"), line(0,0,-1,3),
line(-1,3,-1,0),line(0,0,-1,0) )}}}

The length of the horizontal line is the x-coordinate of (-1,3),
which is -1.  We call the horizontal line x, and label it x=-1

{{{drawing(400,400,-4,4,-4,4, graph(400,400,-4,4,-4,4,(sqrt(.33+x)/sqrt(.33+x))*sqrt(.6-x^2)  ), locate(.3,1.1,B),locate(-1.5,3.4,"(-1,3)"), line(0,0,-1,3),
line(-1,3,-1,0),line(0,0,-1,0), locate(-.95,.4,"x=-1") )}}}

The length of the vertical line is the y-coordinate of (-1,3),
which is 3.  We call the vertical line y, and label it y=3:

{{{drawing(400,400,-4,4,-4,4, graph(400,400,-4,4,-4,4,(sqrt(.33+x)/sqrt(.33+x))*sqrt(.6-x^2)  ), locate(.3,1.1,B),locate(-1.5,3.4,"(-1,3)"), line(0,0,-1,3),
line(-1,3,-1,0),line(0,0,-1,0),locate(-1.6,1.5,"y=3"), locate(-.95,.4,"x=-1") )}}}

So since {{{tan(B) = y/x}}} we can give the
answer 

{{{tan(B) = y/x = 3/(-1) = -3}}}

But if you were asking for the cos(B) or sin(B)
we would need to find the length of the terminal arm,
which is the hypotenuse of the right triangle.  We 
use the Pythagorean theorem:

{{{r^2=x^2+y^2}}}
{{{r^2=(-1)^2+(3)^2}}}
{{{r^2=1+9}}}
{{{r^2=10}}}
{{{r=sqrt(10)}}}

The length of the terminal arm is {{{sqrt(10)}}}.
We call the horizontal line {{{r}}}, and we
label it {{{r=sqrt(10)}}}

{{{drawing(400,400,-4,4,-4,4, graph(400,400,-4,4,-4,4,(sqrt(.33+x)/sqrt(.33+x))*sqrt(.6-x^2)  ), locate(.3,1.1,B),locate(-1.5,3.4,"(-1,3)"), line(0,0,-1,3),
line(-1,3,-1,0),line(0,0,-1,0),locate(-1.6,1.5,"y=3"), locate(-.95,.4,"x=-1"),locate(-.3,1.7,r=sqrt(10)) )}}}

{{{sin(B) = y/r = 3/sqrt(10) = (3sqrt(10))/10}}}

{{{cos(B) = x/r = (-1)/sqrt(10) = (-sqrt(10))/10}}}

Edwin</pre>