Thursday

January 19, 2017
Posted by **Mallory** on Monday, December 20, 2010 at 10:33pm.

- trig -
**Henry**, Tuesday, December 21, 2010 at 12:25pmWhen in STD position, the center i8s at

the origin:

C(h , k),

C(0 , o), P(-3 , 4),

r^2 = (x - h)^2 + (y - k)^2,

r^2 = (-3 - 0)^2 + (4 - 0)^2,

r^2 = 9 + 16 = 25,

r = 5.

rcos(theta) = x -h = -3 -0 = -3,

5cos(theta) = -3,

cos(theta) = -3/5 = -0.6,

theta = 126.9 deg.

Sec(theta) = 1/cos(theta) = 1/-0.6 =

-1.67.