Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

3.1: Sign of an angle

  • Page ID
    23591
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    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.

    Lemma \(\PageIndex{1}\)

    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 \(\alpha = \beta\).