Question 1085571
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We plot the point P(11,-4), draw a line from it to the origin,
so as to make a right triangle with the side opposite &#952; as the
y coordinate of P, and the adjacent side as the x-coordinate
of P.  The red arc is the angle &#952;, and the reference angle
is the one inside the right triangle next to the x-axis. We
use the reference angle to determine the adjacent and opposite
sides. 

{{{drawing(400,8800/47,-2.9,12.9,-5.9,2.9, grid(1),
locate(10,-4,"P(11,-4)"),line(10.94,-.01,10.94,-3.99),
triangle(.01,-.01,10.99,-.01,10.98,-3.98),
red(arc(0,0,2.4,-2.4,0,340)),locate(11.1,-1.85,y=-4),
locate(5.3,.83,x=11),


line(0,0,11,-4) )}}}

The sine is the opposite side over the hypotenuse.
The opposite side of &#952; is the y-ccoordinate of the
point P which is -4.  The adjacent side of &#952; is the
x-coordinate of P which is 11.  

We need to calculate the hypotenuse by the Pythagorean
theorem

{{{hypotenuse^2=adjacent^2+opposite^2}}}
{{{hypotenuse^2=11^2+(-4)^2}}}
{{{hypotenuse^2=121+16}}}
{{{hypotenuse^2=137}}}
{{{hypotenuse=sqrt(137)}}}

So we put {{{sqrt(137)}}} on the hypotenuse:

{{{drawing(400,8800/47,-2.9,12.9,-5.9,2.9, grid(1),
locate(10,-4,"P(11,-4)"),line(10.94,-.01,10.94,-3.99),
triangle(.01,-.01,10.99,-.01,10.98,-3.98),
red(arc(0,0,2.4,-2.4,0,340)),locate(11.1,-1.85,y=-4),
locate(5.3,.83,x=11), locate(5.3,-2.5,sqrt(137)), 


line(0,0,11,-4) )}}}   

Now {{{SINE=OPPOSITE/HYPOTENUSE}}}, so

{{{sin(theta)=(-4)/sqrt(137)=(-4)/11.70469991=-0.3417430631}}}

So if we round to 3 decimal places, the answer is

{{{sin(theta)=-0.342}}} <--the negative sign is very important!

Edwin</pre></b></font>