
# 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.

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 $$\alpha = \beta$$.