# Terminal arm

• June 16th 2009, 03:10 PM
brentwoodbc
Terminal arm
The point (-2,3) lies on the terminal arm of position angle

http://www.quizmebc.ca/images/exams/3990433b.gif

I dont know how to solve this question, I had one where it was simular but gave an equation like y=cotx and I couldnt solve that either.

I know sin=3 and cos=-2 but that is it.
• June 16th 2009, 08:41 PM
yeongil
Quote:

Originally Posted by brentwoodbc
The point (-2,3) lies on the terminal arm of position angle

http://www.quizmebc.ca/images/exams/3990433b.gif

I dont know how to solve this question, I had one where it was simular but gave an equation like y=cotx and I couldnt solve that either.

I still don't know your question. If you're trying to find $\cot \theta$, then the answer would be $-\frac{2}{3}$, because in the cartesian plane, with point (x, y) on the terminal ray, $\cot \theta = \frac{x}{y}$.

Quote:

I know sin=3 and cos=-2 but that is it.
First, watch your notation. You can't write a trig function without an angle. You must write something like
$\sin \theta = 3$ or $\cos \theta = -2$.
Second, neither value is possible. Sine or cosine of any angle will be between -1 and 1 inclusive. If you're using the point (x, y) to figure out sine and cosine, you need these definitions:
$\sin \theta = \frac{y}{r}$ and
$\cos \theta = \frac{x}{r}$.

In other words, you forgot to divide each by r. (r is easy to find, it's just $\sqrt{x^2 + y^2}$).

01
• June 16th 2009, 08:53 PM
brentwoodbc
Quote:

Originally Posted by brentwoodbc
The point (-2,3) lies on the terminal arm of position angle

http://www.quizmebc.ca/images/exams/3990433b.gif

Thats my question how do you solve that.
• June 16th 2009, 08:59 PM
mr fantastic
Quote:

Originally Posted by brentwoodbc
Thats my question how do you solve that.

What you have posted is not clear enough to answer.
• June 16th 2009, 10:05 PM
brentwoodbc
I know that is what I thought but it is like question #31 then that.