# On the unit circle, the point P(-5/13, 12/13) lies on the terminal arm of an angle in standard position?

Jun 18, 2015

The angle is $1.9656$ radians or ${112.62}^{\circ}$
(assuming that is what the question intended to ask).

#### Explanation:

In standard position with the unit length terminal arm at $P \left(- \frac{5}{13} , \frac{12}{13}\right)$
implies
$\textcolor{w h i t e}{\text{XXXX}}$$\cos \left(\theta\right) = - \frac{5}{13}$

which, in turn, implies
$\textcolor{w h i t e}{\text{XXXX}}$$\theta = \arccos \left(- \frac{5}{13}\right)$
which can be solved using a calculator to obtain the result above.

Jun 18, 2015

I have no one sentence answer. See below.

#### Explanation:

If the question was to find the values of the six trigonometric functions, use the definition. Note that
$r = \sqrt{{\left(- \frac{5}{13}\right)}^{2} + {\left(\frac{12}{13}\right)}^{2}} = \sqrt{\frac{25 + 122}{169}} = \sqrt{\frac{169}{169}} = 1$

(The point P is on the unit circle.)

Call the angle $\theta$ (I'll use $r$ in the definitions even though it is $1$.

$\sin \theta = \frac{y}{r} = \frac{y}{1} = \frac{12}{13}$ $\textcolor{w h i t e}{\text{sssss}}$ $\csc \theta = \frac{r}{y} = \frac{1}{y} = \frac{1}{\frac{12}{13}} = \frac{13}{12}$

$\cos \theta = \frac{x}{r} = \frac{x}{1} = - \frac{5}{13}$ $\textcolor{w h i t e}{\text{sss}}$ $\sec \theta = \frac{r}{x} = \frac{1}{x} = \frac{1}{- \frac{5}{13}} = - \frac{13}{5}$

$\tan \theta = \frac{y}{x} = \frac{\frac{12}{13}}{- \frac{5}{13}} = \frac{12}{13} \cdot - \frac{13}{5} = - \frac{12}{5}$

$\cot \theta = \frac{x}{y} = \frac{- \frac{5}{13}}{\frac{12}{13}} = - \frac{5}{13} \cdot \frac{13}{12} = - \frac{5}{12}$