5.1: Right, Acute and Obtuse Angles

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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.

$截屏2021-02-03 下午4.26.12.png$

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.