Calcular: $$ \begin{array}{l} \mathrm{E}=\mathrm{a}+\mathrm{b}+\mathrm{c} \ \mathrm{a}=\left( an 37^{\circ} \cdot \cos 53^{\circ} ight)(5) \ \mathrm{b}=\operatorname{sen} 45^{\circ} \cdot \sec 45^{\circ} \ \mathrm{c}=2 \operatorname{sen} 60^{\circ} \cdot an 30^{\circ}\end{array} $$

Question

Calcular: $$ \begin{array}{l} \mathrm{E}=\mathrm{a}+\mathrm{b}+\mathrm{c} \ \mathrm{a}=\left(	an 37^{\circ} \cdot \cos 53^{\circ}ight)(5) \ \mathrm{b}=\operatorname{sen} 45^{\circ} \cdot \sec 45^{\circ} \ \mathrm{c}=2 \operatorname{sen} 60^{\circ} \cdot 	an 30^{\circ}\end{array} $$

UpStudy AI · Accepted Answer

1. Calcular $$ a $$:

Dado que  
   $$
   a = \left(\tan 37^\circ \cdot \cos 53^\circ\right)(5)
   $$
   
   Se observa que los ángulos $$37^\circ$$ y $$53^\circ$$ son complementarios, ya que  
   $$
   37^\circ + 53^\circ = 90^\circ.
   $$
   
   Además, se utiliza la identidad:  
   $$
   \cos 53^\circ = \sin (90^\circ-53^\circ) = \sin 37^\circ.
   $$
   
   Por lo tanto,  
   $$
   \tan 37^\circ \cdot \cos 53^\circ = \tan 37^\circ \cdot \sin 37^\circ.
   $$
   
   Recordando que  
   $$
   \tan 37^\circ = \frac{\sin 37^\circ}{\cos 37^\circ},
   $$
   
   se tiene  
   $$
   \tan 37^\circ \cdot \sin 37^\circ = \frac{\sin 37^\circ}{\cos 37^\circ} \cdot \sin 37^\circ = \frac{\sin^2 37^\circ}{\cos 37^\circ}.
   $$
   
   Así,  
   $$
   a = 5\cdot \frac{\sin^2 37^\circ}{\cos 37^\circ}.
   $$
   
   Dado que $$37^\circ$$ no es un ángulo notoriamente especial con valor exacto en forma radical, se utilizan valores aproximados:
   
   $$
   \sin 37^\circ \approx 0.6018,\quad \cos 37^\circ \approx 0.7986.
   $$
   
   Entonces,
   
   $$
   \frac{\sin^2 37^\circ}{\cos 37^\circ} \approx \frac{(0.6018)^2}{0.7986} \approx \frac{0.3622}{0.7986} \approx 0.4535.
   $$
   
   Multiplicando por 5:
   
   $$
   a \approx 5 \cdot 0.4535 \approx 2.2675.
   $$
   
2. Calcular $$ b $$:

Se tiene  
   $$
   b = \sin 45^\circ \cdot \sec 45^\circ.
   $$
   
   Sabemos que  
   $$
   \sin 45^\circ = \frac{\sqrt{2}}{2} \quad \text{y} \quad \cos 45^\circ = \frac{\sqrt{2}}{2},
   $$
   
   por lo que  
   $$
   \sec 45^\circ = \frac{1}{\cos 45^\circ} = \frac{2}{\sqrt{2}} = \sqrt{2}.
   $$
   
   Entonces,
   
   $$
   b = \frac{\sqrt{2}}{2} \cdot \sqrt{2} = \frac{2}{2} = 1.
   $$
   
3. Calcular $$ c $$:

Se tiene  
   $$
   c = 2 \sin 60^\circ \cdot \tan 30^\circ.
   $$
   
   Sabemos que  
   $$
   \sin 60^\circ = \frac{\sqrt{3}}{2} \quad \text{y} \quad \tan 30^\circ = \frac{1}{\sqrt{3}}.
   $$
   
   Entonces,
   
   $$
   c = 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}} = 1.
   $$
   
4. Calcular $$ E $$:

Finalmente,  
   $$
   E = a + b + c \approx 2.2675 + 1 + 1 \approx 4.2675.
   $$
   
   Redondeando a dos decimales,  
   $$
   E \approx 4.27.
   $$
$$ E \approx 4.27 $$

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