# How do you sketch the angle whose terminal side in standard position passes through (-7,6sqrt2) and how do you find sin and cos?

Feb 26, 2017

$\sin t = \frac{6 \sqrt{2}}{11}$
$\cos t = - \frac{7}{11}$

#### Explanation:

Consider right triangle MOm, with
Om = x = - 7
mM = y = $6 \sqrt{2}$.
The hypotenuse $O M = \sqrt{49 + 72} = \sqrt{121} = 11$
$\sin t = \frac{6 \sqrt{2}}{11}$
$\cos t = - \frac{7}{11}$