Given point (6,-10) 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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Given point (6,-10) 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 · 2017-03-26T23:38:55

See below

Explanation:

To find the distance to the origin you use the distance formula, $d = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$ $d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)$ .

Plugging in the coordinates of the given point and the origin, $d = \sqrt{{(6 - 0)}^{2} + {(- 10 - 0)}^{2}} = \sqrt{136} = 2 \sqrt{34} \approx 11.7$ $d=sqrt((6-0)^2+("-"10-0)^2)=sqrt136=2sqrt34~~11.7$

To find the angle, get a reference angle using ${tan}^{- 1} (\frac{| y |}{| x |})$ $tan^("-"1)(|y|/|x|)$ and then analyze the point to find the correct quadrant.

${tan}^{- 1} (\frac{10}{6}) \approx 1.03$ $tan^("-"1)(10/6)~~1.03$ and since the point lies on the line below, it is in Quadrant $IV$ $"IV"$ . The value of an angle in Quadrant $IV$ $"IV"$ is $2 π - α$ $2pi-alpha$ where $α$ $alpha$ is the reference angle, so the angle for this point is $5.25$ $5.25$ .

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Given point (6,-10) 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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