# How do you sketch the angle whose terminal side in standard position passes through (2,-sqrt5) and how do you find sin and cos?

Aug 31, 2016

Letting the angle be $\alpha$, we get:
$\sin \alpha = - \frac{\sqrt{5}}{3}$
$\cos \alpha = \frac{2}{3}$

#### Explanation:

Draw a diagram:

Since the angle between the two known sides is right, we can use pythagorean theorem to determine the length of the hypotenuse.

Let the hypotenuse be $x$.

${\left(2\right)}^{2} + {\left(- \sqrt{5}\right)}^{2} = {x}^{2}$

$4 + 5 = {x}^{2}$

${x}^{2} = 9$

$x = \pm 3$

A negative hypotenuse is impossible, so $x = 3$, or the hypotenuse has a measure of $3$ units.

Note that the angle in standard position would be the angle of the triangle opposite the $- \sqrt{5}$.

Let the angle be $\alpha$.

$\sin \alpha = - \frac{\sqrt{5}}{3}$

$\cos \alpha = \frac{2}{3}$

Hopefully this helps!