1. Sean $$ A=\binom{-1}{\sqrt{3}}, B=\binom{2 \sqrt{3}}{-2} $$ y $$ C=\binom{-\sqrt{2} / 2}{-\sqrt{2} / 2} $$. Determine la orientación de las parejas ordenadas $$ (A, B),(B, C) \mathrm{y}(C, A) $$. Halle los valores del Seno y el Coseno para los ángulos $$ \angle(A O B), \angle(B O C) $$ y $$ \angle(C O A) $$.

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Para resolver el problema, analizamos las parejas ordenadas $$ (A, B) $$, $$ (B, C) $$ y $$ (C, A) $$ determinando su orientación y calculando los valores del seno y el coseno de los ángulos $$ \angle AOB $$, $$ \angle BOC $$ y $$ \angle COA $$.

### 1. Orientación de las parejas ordenadas

Para determinar la orientación entre dos vectores $$ \mathbf{u} = (u_x, u_y) $$ y $$ \mathbf{v} = (v_x, v_y) $$, calculamos el determinante:

$$
\text{det}(\mathbf{u}, \mathbf{v}) = u_x v_y - u_y v_x
$$

- **Orientación de $$ (A, B) $$:**

$$
  A = \begin{pmatrix} -1 \ \sqrt{3} \end{pmatrix}, \quad B = \begin{pmatrix} 2\sqrt{3} \ -2 \end{pmatrix}
  $$

$$
  \text{det}(A, B) = (-1)(-2) - (\sqrt{3})(2\sqrt{3}) = 2 - 6 = -4
  $$

**Conclusión:** La orientación es **horaria** (porque el determinante es negativo).

- **Orientación de $$ (B, C) $$:**

$$
  B = \begin{pmatrix} 2\sqrt{3} \ -2 \end{pmatrix}, \quad C = \begin{pmatrix} -\dfrac{\sqrt{2}}{2} \ -\dfrac{\sqrt{2}}{2} \end{pmatrix}
  $$

$$
  \text{det}(B, C) = (2\sqrt{3})\left(-\dfrac{\sqrt{2}}{2}\right) - (-2)\left(-\dfrac{\sqrt{2}}{2}\right) = -\sqrt{6} - \sqrt{2} < 0
  $$

**Conclusión:** La orientación es **horaria**.

- **Orientación de $$ (C, A) $$:**

$$
  C = \begin{pmatrix} -\dfrac{\sqrt{2}}{2} \ -\dfrac{\sqrt{2}}{2} \end{pmatrix}, \quad A = \begin{pmatrix} -1 \ \sqrt{3} \end{pmatrix}
  $$

$$
  \text{det}(C, A) = \left(-\dfrac{\sqrt{2}}{2}\right)(\sqrt{3}) - \left(-\dfrac{\sqrt{2}}{2}\right)(-1) = -\dfrac{\sqrt{6}}{2} - \dfrac{\sqrt{2}}{2} = -\dfrac{\sqrt{6} + \sqrt{2}}{2} < 0
  $$

**Conclusión:** La orientación es **horaria**.

### 2. Cálculo de senos y cosenos de los ángulos

Para dos vectores $$ \mathbf{u} $$ y $$ \mathbf{v} $$, los valores del coseno y seno del ángulo $$ \theta $$ entre ellos se calculan como:

$$
\cos \theta = \dfrac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|}, \quad \sin \theta = \dfrac{|\text{det}(\mathbf{u}, \mathbf{v})|}{|\mathbf{u}| |\mathbf{v}|}
$$

- **Ángulo $$ \angle AOB $$:**

$$
  A = \begin{pmatrix} -1 \ \sqrt{3} \end{pmatrix}, \quad B = \begin{pmatrix} 2\sqrt{3} \ -2 \end{pmatrix}
  $$

$$
  A \cdot B = (-1)(2\sqrt{3}) + (\sqrt{3})(-2) = -4\sqrt{3}
  $$
  
  $$
  |A| = \sqrt{(-1)^2 + (\sqrt{3})^2} = 2, \quad |B| = \sqrt{(2\sqrt{3})^2 + (-2)^2} = 4
  $$

$$
  \cos \angle AOB = \dfrac{-4\sqrt{3}}{2 \times 4} = -\dfrac{\sqrt{3}}{2}
  $$

$$
  \sin \angle AOB = \dfrac{|\text{det}(A, B)|}{2 \times 4} = \dfrac{4}{8} = \dfrac{1}{2}
  $$

- **Ángulo $$ \angle BOC $$:**

$$
  B = \begin{pmatrix} 2\sqrt{3} \ -2 \end{pmatrix}, \quad C = \begin{pmatrix} -\dfrac{\sqrt{2}}{2} \ -\dfrac{\sqrt{2}}{2} \end{pmatrix}
  $$

$$
  B \cdot C = (2\sqrt{3})\left(-\dfrac{\sqrt{2}}{2}\right) + (-2)\left(-\dfrac{\sqrt{2}}{2}\right) = -\sqrt{6} + \sqrt{2}
  $$
  
  $$
  |B| = 4, \quad |C| = 1
  $$

$$
  \cos \angle BOC = \dfrac{-\sqrt{6} + \sqrt{2}}{4} = \dfrac{\sqrt{2} - \sqrt{6}}{4}
  $$

$$
  \sin \angle BOC = \dfrac{|\text{det}(B, C)|}{4 \times 1} = \dfrac{\sqrt{6} + \sqrt{2}}{4}
  $$

- **Ángulo $$ \angle COA $$:**

$$
  C = \begin{pmatrix} -\dfrac{\sqrt{2}}{2} \ -\dfrac{\sqrt{2}}{2} \end{pmatrix}, \quad A = \begin{pmatrix} -1 \ \sqrt{3} \end{pmatrix}
  $$

$$
  C \cdot A = \left(-\dfrac{\sqrt{2}}{2}\right)(-1) + \left(-\dfrac{\sqrt{2}}{2}\right)(\sqrt{3}) = \dfrac{\sqrt{2}}{2} - \dfrac{\sqrt{6}}{2} = \dfrac{\sqrt{2} - \sqrt{6}}{2}
  $$
  
  $$
  |C| = 1, \quad |A| = 2
  $$

$$
  \cos \angle COA = \dfrac{\sqrt{2} - \sqrt{6}}{4}
  $$

$$
  \sin \angle COA = \dfrac{|\text{det}(C, A)|}{1 \times 2} = \dfrac{\sqrt{6} + \sqrt{2}}{4}
  $$

### Resumen de Resultados

- **Orientaciones:**
  - $$ (A, B) $$: **Horaria**
  - $$ (B, C) $$: **Horaria**
  - $$ (C, A) $$: **Horaria**

- **Valores de Seno y Coseno:**
  - **$$ \angle AOB $$:**
    $$
    \cos \angle AOB = -\dfrac{\sqrt{3}}{2}, \quad \sin \angle AOB = \dfrac{1}{2}
    $$
  
  - **$$ \angle BOC $$:**
    $$
    \cos \angle BOC = \dfrac{\sqrt{2} - \sqrt{6}}{4}, \quad \sin \angle BOC = \dfrac{\sqrt{6} + \sqrt{2}}{4}
    $$
  
  - **$$ \angle COA $$:**
    $$
    \cos \angle COA = \dfrac{\sqrt{2} - \sqrt{6}}{4}, \quad \sin \angle COA = \dfrac{\sqrt{6} + \sqrt{2}}{4}
    $$
La orientación de las parejas ordenadas $$ (A, B) $$, $$ (B, C) $$ y $$ (C, A) $$ es horaria. Los valores de seno y coseno para los ángulos $$ \angle AOB $$, $$ \angle BOC $$ y $$ \angle COA $$ son: - **$$ \angle AOB $$:** - $$ \cos \angle AOB = -\dfrac{\sqrt{3}}{2} $$ - $$ \sin \angle AOB = \dfrac{1}{2} $$ - **$$ \angle BOC $$:** - $$ \cos \angle BOC = \dfrac{\sqrt{2} - \sqrt{6}}{4} $$ - $$ \sin \angle BOC = \dfrac{\sqrt{6} + \sqrt{2}}{4} $$ - **$$ \angle COA $$:** - $$ \cos \angle COA = \dfrac{\sqrt{2} - \sqrt{6}}{4} $$ - $$ \sin \angle COA = \dfrac{\sqrt{6} + \sqrt{2}}{4} $$

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