Respuesta:
[tex]\overline {AD}=50sen37\°[/tex]
Explicación paso a paso:
[tex]\frac{sen37\°}{\overline{AD}}=\frac{sen(90\°-23\°-37\°)}{\overline{CD}}\\\\\frac{sen37\°}{\overline{AD}}=\frac{sen30\°}{\overline{CD}}\\\\\overline{AD}=\overline{CD}\frac{sen37\°}{sen30\°}\\\\[/tex]
[tex]\overline {AD}=25\cdot csc30\°\cdot sen37\°\\\\\overline {AD}=50sen37\°\approx30,091[/tex]
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Verified answer
Respuesta:
[tex]\overline {AD}=50sen37\°[/tex]
Explicación paso a paso:
[tex]\frac{sen37\°}{\overline{AD}}=\frac{sen(90\°-23\°-37\°)}{\overline{CD}}\\\\\frac{sen37\°}{\overline{AD}}=\frac{sen30\°}{\overline{CD}}\\\\\overline{AD}=\overline{CD}\frac{sen37\°}{sen30\°}\\\\[/tex]
[tex]\overline {AD}=25\cdot csc30\°\cdot sen37\°\\\\\overline {AD}=50sen37\°\approx30,091[/tex]