You know that
sinØ = y/r
cosØ = x/r
tanØ = y/x
cscØ = r/y
etc
so in any point the first coordinate is the x, the 2nd is the y
1. for (-12, -5) you would be in quadrant III
x = -12, y = -5
use Pythagoras, x^2 + y^2 = r^2 ( where we keep r always positive)
we get r = 13
now simply form the ratios
sinØ = y/r = -5/13
etc
Do the same thing for #2
1. Find csc A, cos A and tan A of an angle A whose terminal side contains (– 12, – 5).
2. Find sin X, sec X and cot X of an angle X whose terminal side contains (– 3, 5).
1 answer