3.1: Sign of an angle
- Page ID
- 23591
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The positive and negative angles can be visualized as counterclockwise and clockwise directions; formally, they are defined the following way:
- The angle \(AOB\) is called positive if \(0 < \measuredangle AOB < \pi\);
- The angle \(AOB\) is called negative if \(\measuredangle AOB < 0\).
Note that according to the above definitions the straight angle as well as the zero angle are neither positive nor negative.
Exercise \(\PageIndex{1}\)
Shoe that \(\angle AOB\) is positive if and only if \(\angle BOA\) is negative.
- Hint
-
Set \(\alpha = \measuredangle AOB\) and \(\beta = \measuredangle BOA\). Note that \(\alpha = \pi\) if and only if \(\beta = \pi\). Otherwise \(\alpha = -\beta\). Hence the result.
Let \(\angle AOB\) be straight. Then \(\angle AOX\) is positive if and only if \(\angle BOX\) is negative.
- Proof
-
Set \(\alpha = \measuredangle AOX\) and \(\beta = \measuredangle BOX\). Since \(\angle AOB\) is straight,
\[\alpha - \beta \equiv \pi.\]
It follows that \(\alpha = \pi \Leftrightarrow \beta = 0\) and \(\alpha = 0 \Leftrightarrow \beta = \pi\). In these two cases the sign of \(\angle AOX\) and \(\angle BOX\) are undefined.
In the remaining cases we have that \(|\alpha| < \pi\) and \(|\beta| < \pi\). If \(\alpha\) and \(\beta\) have the same sign, then \(|\alpha - \beta| < \pi\); the latter contradicts 3.1.1. Hence the statement follows.
Exercise \(\PageIndex{2}\)
Assume that the angles \(ABC\) and \(A'B'C'\) have the same sign and
\(2 \cdot \measuredangle ABC \equiv 2 \cdot \measuredangle A'B'C'.\)
Show that \(\measuredangle ABC = \measuredangle A'B'C'\).
- Hint
-
Set \(\alpha = \measuredangle ABC\), \(\beta = \measuredangle A'B'C'\). Since \(2 \cdot \alpha \equiv 2 \cdot \beta\), Exercise 1.8.1 implies that \(\alpha \equiv \beta\) or \(\alpha \equiv \beta + \pi\). In the latter case the angles have opposite signs which is impossible.
Since \(\alpha, \beta \in (-\pi, \pi]\), equality \(\alpha \equiv \beta\) implies \(\alpha = \beta\).