Hello,

Question

I have a question here that reads: If angle theta lies in quadrant iii & costheta= -2/root29 then determine the exact values of the other 2 trig ratios.

When I take cos-1 of the above I get 111 degrees which would be in quadrant 2... Is that a mistake or am I doing something wrong?

Thanks! :)

Answer 1

nah nah. No inverse functions needed. In fact, you have to keep in mind that the principal values for arccos(x) are 0 to 180, which does not include points in QIII. So, you should have taken arccos(2/√29) = 68°. So, in QIII your angle would be 180+68=248°

Draw the angle, terminating in the point (x,y) where
x = -2
y = -5
r = √29
recall that
sinθ = y/r
cosθ = x/r
tanθ = y/x
Now it's a cinch.