Respuesta:
Explicación paso a paso:
tg A = [tex]\frac{\sqrt{7} }{\sqrt{3} }[/tex] = [tex]\frac{\sqrt{7} }{\sqrt{3} }*\frac{\sqrt{3} }{\sqrt{3} } = \frac{\sqrt{21} }{3}[/tex]
cat. opuesto al ∡A = [tex]\sqrt{21}[/tex]
cat. adyacente al ∡A = 3
hipotenusa h² = ([tex]\sqrt{21}[/tex])² + (3)²
h = [tex]\sqrt{30}[/tex]
Se pide
E = [tex]\sqrt{7.tgB +6.secA}[/tex]
tg B = [tex]\frac{3}{\sqrt{21} }[/tex] = [tex]\frac{3}{\sqrt{21} } *\frac{\sqrt{21} }{\sqrt{21} } =\frac{3\sqrt{21} }{21} = \frac{\sqrt{21} }{7}[/tex]
sec A = [tex]\frac{\sqrt{30} }{3}[/tex]
Reemplaza
E = [tex]\sqrt{7*\frac{\sqrt{21} }{7} +6*\frac{\sqrt{30} }{3} }[/tex]
E = [tex]\sqrt{\sqrt{21} + 2\sqrt{30} }[/tex]
