Respuesta:
[tex]3\sqrt{2}[/tex] ; la correcta es la Opción a )
Explicación paso a paso:
[tex]BC =5[/tex]
[tex]< A = 45[/tex]
[tex]< C = 37[/tex]
Aplicando la ley de los senos:
[tex]\frac{BC}{Sen(<A)} =\frac{AB}{Sen(<C)}[/tex]
[tex]\frac{5}{Sen(45)} = \frac{AB}{Sen(37)}[/tex]
[tex]\frac{5}{\frac{\sqrt{2} }{2} } =\frac{AB}{\frac{3}{5} }[/tex] --------------------- [tex]\frac{10}{\sqrt{2} } =\frac{5AB}{3}[/tex]
[tex](5AB)(\sqrt{2} ) = 3(10)[/tex]
[tex](5\sqrt{2} )(AB ) =30[/tex]
[tex]AB = \frac{30}{5\sqrt{2 } } = \frac{6}{\sqrt{2} }[/tex]
Racionalizando el denominador:
[tex]AB = \frac{6}{\sqrt{2} } *\frac{\sqrt{2} }{\sqrt{2} } =\frac{6\sqrt{2} }{\sqrt{4} } =\frac{6\sqrt{2} }{2} =3\sqrt{2}[/tex]
Luego, [tex]AB = 3\sqrt{2}[/tex]
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Respuesta:
[tex]3\sqrt{2}[/tex] ; la correcta es la Opción a )
Explicación paso a paso:
[tex]BC =5[/tex]
[tex]< A = 45[/tex]
[tex]< C = 37[/tex]
Aplicando la ley de los senos:
[tex]\frac{BC}{Sen(<A)} =\frac{AB}{Sen(<C)}[/tex]
[tex]\frac{5}{Sen(45)} = \frac{AB}{Sen(37)}[/tex]
[tex]\frac{5}{\frac{\sqrt{2} }{2} } =\frac{AB}{\frac{3}{5} }[/tex] --------------------- [tex]\frac{10}{\sqrt{2} } =\frac{5AB}{3}[/tex]
[tex](5AB)(\sqrt{2} ) = 3(10)[/tex]
[tex](5\sqrt{2} )(AB ) =30[/tex]
[tex]AB = \frac{30}{5\sqrt{2 } } = \frac{6}{\sqrt{2} }[/tex]
Racionalizando el denominador:
[tex]AB = \frac{6}{\sqrt{2} } *\frac{\sqrt{2} }{\sqrt{2} } =\frac{6\sqrt{2} }{\sqrt{4} } =\frac{6\sqrt{2} }{2} =3\sqrt{2}[/tex]
Luego, [tex]AB = 3\sqrt{2}[/tex]