posted by kt .
Point P is located at the intersection of a circle with a radius of r and the terminal side of an angle q. Find the coordinates of p to the nearest hundreth.
q = -60° , r = 7
Assuming you are using angle rotation convention, that would put point P into the 4th quadrant.
After drawing the triangle with hypotenuse of 7 and angle of 60º, we can see that
sin -60 = y/7 ----> y = 7(sin -60º) = -6.06
cos -60º = x/7 = 7cos -60º = 3.5
so P is (3.5,-6.06)
we could have used the ratios of sides of the 30-60-90 triangle which are 1:√3:2
then x:y:7 = 1:√3:2
x/1 = 7/2 ---> x = 3.5
y/√3 = 7/2 ---> y = 7√3/2 = 6.06
Realizing where P is located,
P is (3.5,-6.06)