SOLUTION: 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.

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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.
Answer by Edwin McCravy(6938) About Me  (Show Source):
You can put this solution on YOUR website!
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.

drawing%28400%2C400%2C-12%2C12%2C-12%2C12%2C+graph%28400%2C400%2C-12%2C12%2C-12%2C12%2C%28sqrt%28x%2B1%29%2Fsqrt%28x%2B1%29%29%2Asqrt%284-x%5E2%29%29%2C+locate%281.3%2C2.5%2C%27120%B0%27%29%2Clocate%287%2C1.5%2C%27%2810%2C0%29%27%29%2C%0D%0Acircle%280%2C0%2C10%29%2C+line%280%2C0%2C-5%2C5sqrt%283%29%29%2C+locate%28-7%2C9.5%2C%27%28%3F%2C%3F%29%27%29+%29

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%28400%2C400%2C-12%2C12%2C-12%2C12%2C+graph%28400%2C400%2C-12%2C12%2C-12%2C12%2C%28sqrt%28x%2B1%29%2Fsqrt%28x%2B1%29%29%2Asqrt%284-x%5E2%29%2C%28+sqrt%28-x-1%29%2Fsqrt%28-x-1%29%29%2Asqrt%284-x%5E2%29+%29%2C+locate%281.3%2C2.5%2C%27120%B0%27%29%2Clocate%287%2C1.5%2C%27%2810%2C0%29%27%29%2C+locate%28-3%2C1.5%2C%2760%B0%27%29%2C%0D%0Acircle%280%2C0%2C10%29%2C+line%280%2C0%2C-5%2C5sqrt%283%29%29%2Clocate%28-3%2C6%2C10%29%2C+%0D%0Alocate%28-7%2C9.5%2C%27%28%3F%2C%3F%29%27%29+%29

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

drawing%28400%2C400%2C-12%2C12%2C-12%2C12%2C+graph%28400%2C400%2C-12%2C12%2C-12%2C12%2C%28sqrt%28x%2B1%29%2Fsqrt%28x%2B1%29%29%2Asqrt%284-x%5E2%29%2C%28+sqrt%28-x-1%29%2Fsqrt%28-x-1%29%29%2Asqrt%284-x%5E2%29+%29%2C+locate%281.3%2C2.5%2C%27120%B0%27%29%2Clocate%287%2C1.5%2C%27%2810%2C0%29%27%29%2C+locate%28-3%2C1.5%2C%2760%B0%27%29%2C%0D%0Acircle%280%2C0%2C10%29%2C+line%280%2C0%2C-5%2C5sqrt%283%29%29%2Clocate%28-3%2C6%2C10%29%2C+%0D%0Alocate%28-7%2C9.5%2C%27%28%3F%2C%3F%29%27%29%2Cline%28-5%2C0%2C-5%2C5%2Asqrt%283%29%29+%29

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%28400%2C400%2C-12%2C12%2C-12%2C12%2C+graph%28400%2C400%2C-12%2C12%2C-12%2C12%2C%28sqrt%28x%2B1%29%2Fsqrt%28x%2B1%29%29%2Asqrt%284-x%5E2%29%2C%28+sqrt%28-x-1%29%2Fsqrt%28-x-1%29%29%2Asqrt%284-x%5E2%29+%29%2C+locate%281.3%2C2.5%2C%27120%B0%27%29%2Clocate%287%2C1.5%2C%27%2810%2C0%29%27%29%2C+locate%28-3%2C1.5%2C%2760%B0%27%29%2C%0D%0Acircle%280%2C0%2C10%29%2C+line%280%2C0%2C-5%2C5sqrt%283%29%29%2Clocate%28-3%2C6%2C10%29%2Clocate%28-3%2C-1%2C5%29%2C+%0D%0Alocate%28-7%2C9.5%2C%27%28-5%2C%3F%29%27%29%2Cline%28-5%2C0%2C-5%2C5%2Asqrt%283%29%29+%29



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

a%5E2%2Bb%5E2=c%5E2

5%5E2%2Bb%5E2=10%5E2

25%2Bb%5E2=100

b%5E2=100-25

b%5E2=75

b=sqrt%2875%29

b=sqrt%2825%2A3%29

b=5sqrt%283%29 

drawing%28400%2C400%2C-12%2C12%2C-12%2C12%2C+graph%28400%2C400%2C-12%2C12%2C-12%2C12%2C%28sqrt%28x%2B1%29%2Fsqrt%28x%2B1%29%29%2Asqrt%284-x%5E2%29%2C%28+sqrt%28-x-1%29%2Fsqrt%28-x-1%29%29%2Asqrt%284-x%5E2%29+%29%2C+locate%281.3%2C2.5%2C%27120%B0%27%29%2Clocate%287%2C1.5%2C%27%2810%2C0%29%27%29%2C+locate%28-3%2C1.5%2C%2760%B0%27%29%2C%0D%0Acircle%280%2C0%2C10%29%2C+line%280%2C0%2C-5%2C5sqrt%283%29%29%2Clocate%28-3%2C6%2C10%29%2Clocate%28-3%2C-1%2C5%29%2C+%0D%0Alocate%28-7%2C9.5%2C%27%28-5%2C5%27%29%2Cline%28-5%2C0%2C-5%2C5sqrt%283%29%29%2C%0D%0Alocate%28-7%2C5%2C5sqrt%283%29%29%2Clocate%28-4.5%2C9.5%2C%27V3%29%27%29%0D%0A%0D%0A+%29

So the coordinates of the point is (-5,5sqrt%283%29).

Edwin