
# 5.1: Right, Acute and Obtuse Angles


• If $$|\measuredangle AOB| = \dfrac{\pi}{2}$$, we say that $$\angle AOB$$ is right;
• If $$|\measuredangle AOB| < \dfrac{\pi}{2}$$, we say that $$\angle AOB$$ is acute;
• If $$|\measuredangle AOB| > \dfrac{\pi}{2}$$, we say that $$\angle AOB$$ is obtuse.

On the diagrams, the right angles will be marked with a little square, as shown.

If $$\angle AOB$$ is right, we say also that $$[OA)$$ is perpendicular to $$[OB)$$; it will be written as $$[OA) \perp [OB)$$. From Theorem 2.4.1, it follows that two lines $$(OA)$$ and $$(OB)$$ are appropriately called perpendicular, if $$[OA) \perp [OB)$$. In this case we also write $$(OA) \perp (OB)$$.

Exercise $$\PageIndex{1}$$

Assume point $$O$$ lies between $$A$$ and $$B$$ and $$X \ne O$$. Show that $$\angle XOA$$ is acute if and only if $$\angle XOB$$ is obtuse.

Hint

By Axiom IIIb and Theorem 2.4.1, we have $$\measuredangle XOA - \measuredangle XOB \equiv \pi$$. Since $$|\measuredangle XOA|, |\measuredangle XOB| \le \pi$$, we get that $$|\measuredangle XOA| + |\measuredangle XOB| = \pi$$. Hence the statement follows.