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Revision difference : terminal ray |
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| Let an angle whose \PMlinkescapetext{measure} in radians is $\theta$ be placed \PMlinkescapetext{onto} the Cartesian plane such that one of its rays $R_1$ corresponds to the nonnegative $x$ axis and one can go from the point $(1,0)$ to the point that is the intersection of the other ray $R_2$ of the angle with the circle $x^2+y^2=1$ by traveling exactly $\theta$ units on the circle. (If $\theta$ is positive, the distance should be traveled counterclockwise; otherwise, the distance $|\theta|$ should be traveled clockwise. Also, note that ``other ray'' is used quite loosely, as it may also correspond to the positive $x$ axis also.) Then $R_2$ is the \emph{terminal ray} of the angle. |
Let an angle whose \PMlinkescapetext{measure} in radians is $\theta$ be placed \PMlinkescapetext{onto} the Cartesian plane such that one of its rays $R_1$ corresponds to the nonnegative $x$ axis and one can go from the point $(1,0)$ to the point that is the intersection of the other ray $R_2$ of the angle with the circle $x^2+y^2=1$ by traveling exactly $\theta$ units on the circle. (If $\theta$ is positive, the distance should be traveled counterclockwise; otherwise, the distance $|\theta|$ should be traveled clockwise. Also, note that ``other ray'' is used quite loosely, as it may also correspond to the positive $x$ axis also.) Then $R_2$ is the \emph{terminal ray} of the angle. |
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| The picture below shows the terminal ray $R_2$ of the angle $\displaystyle \theta=\frac{2\pi}{3}$. |
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| \begin{center} |
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| \begin{pspicture}(-2,-2)(2,2) |
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| \psset{unit=0.8cm} |
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| \psline{<->}(-2,0)(2,0) |
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| \rput[b](2,0){$x$} |
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| \psline{<->}(0,-2)(0,2) |
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| \rput[l](0,2){$y$} |
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| \psline[linewidth=1.5pt]{->}(0,0)(2,0) |
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| \rput[b](0.5,0){$R_1$} |
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| \psline[linewidth=1.5pt]{->}(0,0)(0.866,-0.5) |
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| \rput[b](1.366,-0.5){$R_2$} |
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| \psarc(0,0){0.3}{0}{120} |
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| \rput[b](0.3,0.3){$\theta$} |
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| \end{pspicture} |
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| \end{center} |
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