Bradley Carroll
11/03/2023 · Middle School

1 De un triángulo sabemos que: \( \mathrm{a}=6 \mathrm{~m}, \mathrm{~B}=45^{\circ} \) y \( \mathrm{C}=105^{\circ} \). Calcula los restantes elementos.

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Para resolver el triángulo dado, utilizaremos la ley de los senos y la suma de los ángulos de un triángulo. 1. **Encontrar el ángulo A**: Sabemos que la suma de los ángulos de un triángulo es \(180^\circ\). Por lo tanto, podemos calcular el ángulo A: \[ A = 180^\circ - B - C = 180^\circ - 45^\circ - 105^\circ = 30^\circ \] 2. **Encontrar el lado b**: Usamos la ley de los senos, que establece que: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \] Primero, calculamos \(b\): \[ \frac{a}{\sin A} = \frac{b}{\sin B} \] Sustituyendo los valores que tenemos: \[ \frac{6 \, \text{m}}{\sin(30^\circ)} = \frac{b}{\sin(45^\circ)} \] Sabemos que \(\sin(30^\circ) = 0.5\) y \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\): \[ \frac{6 \, \text{m}}{0.5} = \frac{b}{\frac{\sqrt{2}}{2}} \] Simplificando: \[ 12 \, \text{m} = \frac{b}{\frac{\sqrt{2}}{2}} \] Multiplicamos ambos lados por \(\frac{\sqrt{2}}{2}\): \[ b = 12 \, \text{m} \cdot \frac{\sqrt{2}}{2} = 6\sqrt{2} \, \text{m} \approx 8.49 \, \text{m} \] 3. **Encontrar el lado c**: Ahora calculamos \(c\) usando la ley de los senos nuevamente: \[ \frac{a}{\sin A} = \frac{c}{\sin C} \] Sustituyendo los valores: \[ \frac{6 \, \text{m}}{\sin(30^\circ)} = \frac{c}{\sin(105^\circ)} \] Sabemos que \(\sin(105^\circ) = \sin(180^\circ - 75^\circ) = \sin(75^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4}\): \[ \frac{6 \, \text{m}}{0.5} = \frac{c}{\frac{\sqrt{6} + \sqrt{2}}{4}} \] Simplificando: \[ 12 \, \text{m} = \frac{c}{\frac{\sqrt{6} + \sqrt{2}}{4}} \] Multiplicamos ambos lados por \(\frac{\sqrt{6} + \sqrt{2}}{4}\): \[ c = 12 \, \text{m} \cdot \frac{\sqrt{6} + \sqrt{2}}{4} = 3(\sqrt{6} + \sqrt{2}) \, \text{m} \approx 10.39 \, \text{m} \] 4. **Resumen de los elementos del triángulo**: - \(A = 30^\circ\) - \(B = 45^\circ\) - \(C = 105^\circ\) - \(a = 6 \, \text{m}\) - \(b \approx 8.49 \, \text{m}\) - \(c \approx 10.39 \, \text{m}\) Por lo tanto, los elementos restantes del triángulo son: - \(A = 30^\circ\) - \(b \approx 8.49 \, \text{m}\) - \(c \approx 10.39 \, \text{m}\)

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- \(A = 30^\circ\) - \(b \approx 8.49 \, \text{m}\) - \(c \approx 10.39 \, \text{m}\)
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