Question 141710
A circle centered at the origin has a radius of 10.If the point(10,0)is rotated 120 degrees counter-clockwise, determine the coordinates of the new point in exact radical form.
<pre><font size = 4 color = "indigo"><b>
{{{drawing(400,400,-12,12,-12,12, graph(400,400,-12,12,-12,12,(sqrt(x+1)/sqrt(x+1))*sqrt(4-x^2)), locate(1.3,2.5,"120°"),locate(7,1.5,"(10,0)"),
circle(0,0,10), line(0,0,-5,5sqrt(3)), locate(-7,9.5,"(?,?)") )}}}

Now we subtract 180°-120° and find that the angle between the slanted
line and the left side of the x-axis is 60° (indicated by the green
curved line, so we label it 60°:

{{{drawing(400,400,-12,12,-12,12, graph(400,400,-12,12,-12,12,(sqrt(x+1)/sqrt(x+1))*sqrt(4-x^2),( sqrt(-x-1)/sqrt(-x-1))*sqrt(4-x^2) ), locate(1.3,2.5,"120°"),locate(7,1.5,"(10,0)"), locate(-3,1.5,"60°"),
circle(0,0,10), line(0,0,-5,5sqrt(3)),locate(-3,6,10), 
locate(-7,9.5,"(?,?)") )}}}

Now from the point in question we draw a perpendicular down to the
x-axis:

{{{drawing(400,400,-12,12,-12,12, graph(400,400,-12,12,-12,12,(sqrt(x+1)/sqrt(x+1))*sqrt(4-x^2),( sqrt(-x-1)/sqrt(-x-1))*sqrt(4-x^2) ), locate(1.3,2.5,"120°"),locate(7,1.5,"(10,0)"), locate(-3,1.5,"60°"),
circle(0,0,10), line(0,0,-5,5sqrt(3)),locate(-3,6,10), 
locate(-7,9.5,"(?,?)"),line(-5,0,-5,5*sqrt(3)) )}}}

This forms a special right triangle.  

We are supposed to know that if a right triangle has an acute angle 
of 60°, then one of the sides of the 60° angle is one-half the other.
So since the hypotenuse is 10, its bottom leg is 5.

That tells us that the x-coordinate of the point in question is {{{-5}}}.

{{{drawing(400,400,-12,12,-12,12, graph(400,400,-12,12,-12,12,(sqrt(x+1)/sqrt(x+1))*sqrt(4-x^2),( sqrt(-x-1)/sqrt(-x-1))*sqrt(4-x^2) ), locate(1.3,2.5,"120°"),locate(7,1.5,"(10,0)"), locate(-3,1.5,"60°"),
circle(0,0,10), line(0,0,-5,5sqrt(3)),locate(-3,6,10),locate(-3,-1,5), 
locate(-7,9.5,"(-5,?)"),line(-5,0,-5,5*sqrt(3)) )}}}



We can calculate the vertical side of that right triangle by
the Pythagorean theorem:

{{{a^2+b^2=c^2}}}

{{{5^2+b^2=10^2}}}

{{{25+b^2=100}}}

{{{b^2=100-25}}}

{{{b^2=75}}}

{{{b=sqrt(75)}}}

{{{b=sqrt(25*3)}}}

{{{b=5sqrt(3)}}} 

{{{drawing(400,400,-12,12,-12,12, graph(400,400,-12,12,-12,12,(sqrt(x+1)/sqrt(x+1))*sqrt(4-x^2),( sqrt(-x-1)/sqrt(-x-1))*sqrt(4-x^2) ), locate(1.3,2.5,"120°"),locate(7,1.5,"(10,0)"), locate(-3,1.5,"60°"),
circle(0,0,10), line(0,0,-5,5sqrt(3)),locate(-3,6,10),locate(-3,-1,5), 
locate(-7,9.5,"(-5,5"),line(-5,0,-5,5sqrt(3)),
locate(-7,5,5sqrt(3)),locate(-4.5,9.5,"V3)")

 )}}}

So the coordinates of the point is ({{{-5}}},{{{5sqrt(3)}}}).

Edwin</pre>