3.1: Signo de un ángulo
- Page ID
- 114414
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Los ángulos positivo y negativo se pueden visualizar en sentido antihorario y en sentido horario; formalmente, se definen de la siguiente manera:
- El ángulo\(AOB\) se llama positivo si\(0 < \measuredangle AOB < \pi\);
- El ángulo\(AOB\) se llama negativo si\(\measuredangle AOB < 0\).
Obsérvese que de acuerdo con las definiciones anteriores el ángulo recto así como el ángulo cero no son ni positivos ni negativos.
Ejercicio\(\PageIndex{1}\)
Zapato que\(\angle AOB\) es positivo si y solo si\(\angle BOA\) es negativo.
- Pista
-
Establecer\(\alpha = \measuredangle AOB\) y\(\beta = \measuredangle BOA\). Tenga en cuenta que\(\alpha = \pi\) si y solo si\(\beta = \pi\). De lo contrario\(\alpha = -\beta\). De ahí el resultado.
Dejar\(\angle AOB\) ser recto. Entonces\(\angle AOX\) es positivo si y sólo si\(\angle BOX\) es negativo.
- Prueba
-
Establecer\(\alpha = \measuredangle AOX\) y\(\beta = \measuredangle BOX\). Dado que\(\angle AOB\) es recto,
\[\alpha - \beta \equiv \pi.\]
De ello se deduce que\(\alpha = \pi \Leftrightarrow \beta = 0\) y\(\alpha = 0 \Leftrightarrow \beta = \pi\). En estos dos casos el signo de\(\angle AOX\) y\(\angle BOX\) son indefinidos.
En los casos restantes tenemos eso\(|\alpha| < \pi\) y\(|\beta| < \pi\). Si\(\alpha\) y\(\beta\) tienen el mismo signo, entonces\(|\alpha - \beta| < \pi\); este último contradice 3.1.1. De ahí sigue el enunciado.
Ejercicio\(\PageIndex{2}\)
Supongamos que los ángulos\(ABC\) y\(A'B'C'\) tienen el mismo signo y
\(2 \cdot \measuredangle ABC \equiv 2 \cdot \measuredangle A'B'C'.\)
\(\measuredangle ABC = \measuredangle A'B'C'\)Demuéstralo.
- Pista
-
Establecer\(\alpha = \measuredangle ABC\),\(\beta = \measuredangle A'B'C'\). Ya que\(2 \cdot \alpha \equiv 2 \cdot \beta\), Ejercicio 1.8.1 implica que\(\alpha \equiv \beta\) o\(\alpha \equiv \beta + \pi\). En este último caso los ángulos tienen signos opuestos lo cual es imposible.
Ya que\(\alpha, \beta \in (-\pi, \pi]\), la igualdad\(\alpha \equiv \beta\) implica\(\alpha = \beta\).