trig
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)
or
we could have used the ratios of sides of the 306090 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)
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