DOY CORONA SI LO RESUELVES

[tex]\large\textsc{Tri\'angulo de 53 y 37}}[/tex]
[tex]\large\textsc{Razones trigonom\'etricas}}[/tex]
[tex]\text{Estas ser\'an las que utilizaremos}[/tex]
[tex]\mathbf{Sen(x)=\dfrac{Cateto\:opuesto}{Hipotenusa} ~~~~~~~~~~~\mathbf{Tan(x)=\dfrac{Cateto\:opuesto}{Cateto\:adyacente} }}[/tex]
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Primero calculamos el seno de 53° y la tangente de 37°
[tex]\mathrm{\sin=\dfrac{CO}{H} }[/tex]
[tex]\mathrm{\sin(53^o)=\dfrac{4k}{5k} }[/tex]
[tex]\boxed{\mathrm{\sin(53^o)=\dfrac{4}{5} }}[/tex]
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[tex]\mathrm{\tan=\dfrac{CO}{CA} }[/tex]
[tex]\mathrm{\tan(37^o)=\dfrac{3k}{4k} }[/tex]
[tex]\boxed{\mathrm{\tan(37^o)=\dfrac{3}{4} }}[/tex]
Y hallamos M
[tex]\mathrm{M = sen(53^o)+tg(37^o)}[/tex]
[tex]\mathrm{M = \dfrac{4}{5} + \dfrac{3}{4} }[/tex]
[tex]\mathrm{M = \dfrac{(4 \times 4)+(5 \times 3)}{5 \times 4}}[/tex]
[tex]\mathrm{M = \dfrac{16+15}{20}}[/tex]
[tex]\large\boxed{\mathbf{M = \dfrac{31}{20}}}[/tex]
RPTA: M = 31/20
Espero que te sirva! ^^
Saludos, Math_and_fisic_girl