Vaughan Ramos
04/26/2024 · Junior High School
Razonamiento Resuelve cada triángulo \( A B C \) sabiendo que \( \Varangle C \) es un ángulo recto, a. \( \Varangle A=55^{\circ} \) y \( a=36 \mathrm{~cm} \) b. \( c=20 \mathrm{~cm} \) y \( b=12 \mathrm{~cm} \) c. \( a=9 \mathrm{~cm} \) y \( b=7,5 \mathrm{~cm} \)
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Para resolver cada uno de los triángulos \( ABC \) donde \( \angle C \) es un ángulo recto, utilizaremos las propiedades de los triángulos rectángulos y el teorema de Pitágoras, así como las razones trigonométricas.
### a. \( \angle A = 55^\circ \) y \( a = 36 \, \text{cm} \)
1. **Encontrar \( \angle B \)**:
\[
\angle B = 90^\circ - \angle A = 90^\circ - 55^\circ = 35^\circ
\]
2. **Encontrar \( b \) usando la función seno**:
\[
\sin(\angle A) = \frac{a}{c} \implies c = \frac{a}{\sin(\angle A)} = \frac{36}{\sin(55^\circ)}
\]
Calculando \( \sin(55^\circ) \):
\[
\sin(55^\circ) \approx 0.8192 \implies c \approx \frac{36}{0.8192} \approx 43.9 \, \text{cm}
\]
3. **Encontrar \( b \) usando la función coseno**:
\[
\cos(\angle A) = \frac{b}{c} \implies b = c \cdot \cos(\angle A) = 43.9 \cdot \cos(55^\circ)
\]
Calculando \( \cos(55^\circ) \):
\[
\cos(55^\circ) \approx 0.5736 \implies b \approx 43.9 \cdot 0.5736 \approx 25.2 \, \text{cm}
\]
**Resultados**:
- \( \angle A = 55^\circ \)
- \( \angle B = 35^\circ \)
- \( a = 36 \, \text{cm} \)
- \( b \approx 25.2 \, \text{cm} \)
- \( c \approx 43.9 \, \text{cm} \)
---
### b. \( c = 20 \, \text{cm} \) y \( b = 12 \, \text{cm} \)
1. **Encontrar \( a \) usando el teorema de Pitágoras**:
\[
a^2 + b^2 = c^2 \implies a^2 + 12^2 = 20^2 \implies a^2 + 144 = 400 \implies a^2 = 256 \implies a = 16 \, \text{cm}
\]
2. **Encontrar \( \angle A \) usando la función seno**:
\[
\sin(\angle A) = \frac{a}{c} = \frac{16}{20} = 0.8 \implies \angle A = \arcsin(0.8) \approx 53.13^\circ
\]
3. **Encontrar \( \angle B \)**:
\[
\angle B = 90^\circ - \angle A \approx 90^\circ - 53.13^\circ \approx 36.87^\circ
\]
**Resultados**:
- \( \angle A \approx 53.13^\circ \)
- \( \angle B \approx 36.87^\circ \)
- \( a = 16 \, \text{cm} \)
- \( b = 12 \, \text{cm} \)
- \( c = 20 \, \text{cm} \)
---
### c. \( a = 9 \, \text{cm} \) y \( b = 7.5 \, \text{cm} \)
1. **Encontrar \( c \) usando el teorema de Pitágoras**:
\[
c^2 = a^2 + b^2 \implies c^2 = 9^2 + 7.5^2 = 81 + 56.25 = 137.25 \implies c = \sqrt{137.25} \approx 11.7 \, \text{cm}
\]
2. **Encontrar \( \angle A \) usando la función seno**:
\[
\sin(\angle A) = \frac{a}{c} = \frac{9}{11.7} \approx 0.7692 \implies \angle A \approx \arcsin(0.769
Quick Answer
### a. \( \angle A = 55^\circ \), \( b \approx 25.2 \, \text{cm} \), \( c \approx 43.9 \, \text{cm} \)
### b. \( \angle A \approx 53.13^\circ \), \( \angle B \approx 36.87^\circ \), \( a = 16 \, \text{cm} \), \( b = 12 \, \text{cm} \), \( c = 20 \, \text{cm} \)
### c. \( \angle A \approx 53.13^\circ \), \( \angle B \approx 36.87^\circ \), \( c \approx 11.7 \, \text{cm} \), \( a = 9 \, \text{cm} \), \( b = 7.5 \, \text{cm} \)
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