# Terminal Arm Location

• Aug 28th 2012, 03:41 PM
waleedrabbani
Terminal Arm Location
You are given

Cos(theta) = -(5/13) .0<theta<2pi

In which quadrant(s) can the terminal arm be,.

I believe quadrants 2 and 4 but the back of the book is saying 2, and 3 (Angry)

I think it is 2 and 4 because the points of its location could be (-5, 13) Q2 or (5, -13) Q4

If i am wrong how do i solve it?

THANKS ALOT,
• Aug 28th 2012, 03:53 PM
emakarov
Re: Terminal Arm Location
Quote:

Originally Posted by waleedrabbani
I think it is 2 and 4 because the points of its location could be (-5, 13) Q2 or (5, -13) Q4

So, cosine is the x-coordinate. Do both of these points have negative x-coordinates?
• Aug 28th 2012, 04:33 PM
waleedrabbani
Re: Terminal Arm Location
I still dont get it what points would i use?

(-5, 13) and (-5, -13)

13 is the hypotenouse it cant be negative can it?
• Aug 28th 2012, 04:46 PM
emakarov
Re: Terminal Arm Location
The two possible points are (-5, 12) and (-5, -12).
• Aug 28th 2012, 05:00 PM
waleedrabbani
Re: Terminal Arm Location
Quote:

Originally Posted by emakarov
The two possible points are (-5, 12) and (-5, -12).

Oh i see,

Thanks
• Aug 28th 2012, 05:30 PM
Plato
Re: Terminal Arm Location
Quote:

Originally Posted by waleedrabbani
You are given Cos(theta) = -(5/13) .0<theta<2pi
In which quadrant(s) can the terminal arm be,.

Well $cos(\theta)=-\frac{5}{13}$ means that $x=-5~\&~y=12\text{ or }x=5~\&~y=-12$.
That means $\theta\in II\text{ or }III.$