# Math Help - Value of function given point?

1. ## Value of function given point?

No idea where to even start on this one. I'm asked to find the exact value of $\sec \theta$ for angle $\theta$ in standard position if the point $(-3, 2)$ lies on its terminal side. I'm horrible with standard positions and the terminal side, can anyone help shine some light on this problem?

Thanks!

2. Originally Posted by Failbait
No idea where to even start on this one. I'm asked to find the exact value of $\sec \theta$ for angle $\theta$ in standard position if the point $(-3, 2)$ lies on its terminal side. I'm horrible with standard positions and the terminal side, can anyone help shine some light on this problem?

Thanks!
just note that the triangle is in the second quadrant. so $\sec \theta$ is negative.

you have a right-triangle with an acute angle $\theta$. opposite to that angle is a side of length 2, and adjacent to that angle is a side of size 3.

what is the hypotenuse of this triangle?

when we find the hypotenuse, how do we get $\sec \theta$?

3. (-3,2) is a point in the second quadrant.
Both the cosine and the secant functions are negative in the second quadrant.
$\begin{array}{l}
\phi \equiv \left( {x,y} \right),\;r = \sqrt {x^2 + y^2 } \\
\cos (\phi ) = \frac{x}{r}\quad \& \quad \sec (\phi ) = \frac{1}{{\cos (\phi )}} \\
\end{array}$

4. Alright, using the Pythagorean theorem, I've got that the hypotenuse is $\sqrt{13}$. Given that, would $\sec \theta = -(\sqrt{13}/3)$?

Thanks again.

5. Originally Posted by Failbait
Alright, using the Pythagorean theorem, I've got that the hypotenuse is $\sqrt{13}$. Given that, would $\sec \theta = -(\sqrt{13}/3)$?

Thanks again.
yes