Question 23293: Point P is the intersection of the terminal arm of angle Q in standard position and the unit circle with centre (0,0). If P is in quadrant 3 and cosQ = m, determine the coordinates of P in terms of m.
A. (-m, sqrt(1-m^2))
A. (-m, -sqrt(1-m^2))
A. (m, sqrt(1-m^2))
A. (m, -sqrt(1-m^2))
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Draw the picture.
The point is "m" to the left of the y-axis.
Now cosQ =m =m/1. The point is "1" away
from the origin because it is on a unit-
circle. Using Pythagoras you get
1^2 = m^2 + ?^2.
The ? = sqrt(1-m^2) but it is negative because
it is a Y-value in the III quadrant.
P is the point (m,-sqrt(1+m^2).
Cheers,
Stan H.
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