# Given point (-8,8) how do you find the distance of the point from the origin, then find the measure of the angle in standard position whose terminal side contains the point?

Mar 19, 2018

Measure of the angle $\theta = {135}^{\circ}$ or ${\left(\frac{3 \pi}{4}\right)}^{c}$

#### Explanation:

Distance formula : $d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

Given point $A \left(- 8 , 8\right)$ and Origin $O \left(0 , 0\right)$

$\vec{O A} = r = \sqrt{{\left(- 8\right)}^{2} + {8}^{2}} = \sqrt{128} = 8 \sqrt{2}$

$x = r \cos \theta$, $y = r \sin \theta$

$- 8 = 8 \sqrt{2} \cos \theta$

$\cos \theta = - \frac{\cancel{8}}{\cancel{8} \sqrt{2}} = - \frac{1}{\sqrt{2}}$ or theta =color(purple)( - pi/4 = (pi - pi/4) = (3pi)/4

Similarly, $8 \sqrt{2} \sin \theta = 8$

$\sin \theta = \frac{8}{8 \sqrt{2}} = \frac{1}{\sqrt{2}}$ or $\theta = \frac{\pi}{4}$ or = pi - pi/4) = (3pi)/4

Slope of the line $m = \tan \theta = - \frac{8}{8} = - 1$

$x$ $\left(-\right)$ and $y$ $\left(+\right)$ in II quadrant.

Hence $\theta = {\tan}^{-} 1 \left(- \frac{8}{8}\right) = {\tan}^{-} 1 - 1 = - 1 = \frac{3 \pi}{4}$

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