3.1: Sign of an angle
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 \(\alpha = \beta\).