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?

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Answer 1 · 2018-03-19T08:19:56

Measure of the angle $theta = 135^@$ or $((3pi)/4)^c$

Distance formula : $d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)$

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

$vec(OA) = r = sqrt ((-8)^2 + 8^2) = sqrt 128 = 8sqrt2$

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

$- 8 = 8 sqrt2 cos theta$

$cos theta = -cancel8 / (cancel8 sqrt2) = -1/sqrt2$ or $theta =color(purple)( - pi/4 = (pi - pi/4) = (3pi)/4$

Similarly, $8 sqrt2 sin theta = 8$

$sin theta = 8 / (8 sqrt2) = 1/sqrt2$ or $theta = pi/4$ or = $pi - pi/4) = (3pi)/4$

Slope of the line $m = tan theta = -8 / 8 = -1$

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

Hence $theta = tan ^-1 (-8/8) = tan^-1 -1 = -1 = (3pi)/4$

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