Galera me ajuda ai por favor
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Olá Gizelle
Vamos calcular o (x) usando o lei senos , veja no anexo baixo
![\frac{x}{sen80 ^{0} } = \frac{80}{sen65^{o} } --\ \textgreater \ valor\ aprox\ de\ [ sen80 =0,98\ e \ sen65=0,91] \\ \\ \frac{x}{0,98} = \frac{80}{0,91}--\ \textgreater \ isolando(x)\ temos. \\ \\ x= \frac{80.(0,98)}{0,91} \\ \\ x=86,1538...--\ \textgreater \ aproximando\ fica \\ \\ \boxed{x=86,2m}](https://tex.z-dn.net/?f=+%5Cfrac%7Bx%7D%7Bsen80+%5E%7B0%7D+%7D+%3D+%5Cfrac%7B80%7D%7Bsen65%5E%7Bo%7D+%7D+--%5C+%5Ctextgreater+%5C+valor%5C+aprox%5C+de%5C+%5B+sen80+%3D0%2C98%5C+e+%5C+sen65%3D0%2C91%5D+%5C%5C++%5C%5C++%5Cfrac%7Bx%7D%7B0%2C98%7D+%3D+%5Cfrac%7B80%7D%7B0%2C91%7D--%5C+%5Ctextgreater+%5C+isolando%28x%29%5C+temos.+%5C%5C++%5C%5C+x%3D+%5Cfrac%7B80.%280%2C98%29%7D%7B0%2C91%7D+++%5C%5C++%5C%5C+x%3D86%2C1538...--%5C+%5Ctextgreater+%5C+aproximando%5C+fica+%5C%5C++%5C%5C+%5Cboxed%7Bx%3D86%2C2m%7D "\frac{x}{sen80 ^{0} } = \frac{80}{sen65^{o} } --\ \textgreater \ valor\ aprox\ de\ [ sen80 =0,98\ e \ sen65=0,91] \\ \\ \frac{x}{0,98} = \frac{80}{0,91}--\ \textgreater \ isolando(x)\ temos. \\ \\ x= \frac{80.(0,98)}{0,91} \\ \\ x=86,1538...--\ \textgreater \ aproximando\ fica \\ \\ \boxed{x=86,2m}")
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Já sabendo o valor de (x) calculamos o ângulo (β), resolvemos por lei de seno tambem, assim.
![\frac{x}{sen \beta } = \frac{82}{sen55^{o} } --\ \textgreater \ valor \ aproximado\ de[ sen55=0,82]\ e\ [ x=86,2] \\ \\ \frac{86,2}{sen \beta } = \frac{82}{0,82} ---\ \textgreater \ isolamos\ (sen \beta ), fica. \\ \\ sen \beta = \frac{86,2.(0,82)}{82} \\ \\ sen \beta =0,862--\ \textgreater \ por\ arcosen \ de\ \beta ,temos. \\ \\ \beta = sen^{-1} 0,862 \\ \\ \beta =59,5418.....---\ \textgreater \ aproximado \\ \\ \boxed{ \boxed{\beta = 59,5^{o} }}](https://tex.z-dn.net/?f=+%5Cfrac%7Bx%7D%7Bsen+%5Cbeta+%7D+%3D+%5Cfrac%7B82%7D%7Bsen55%5E%7Bo%7D+%7D+--%5C+%5Ctextgreater+%5C+valor+%5C+aproximado%5C+de%5B+sen55%3D0%2C82%5D%5C+e%5C+%5B+x%3D86%2C2%5D+%5C%5C++%5C%5C++%5Cfrac%7B86%2C2%7D%7Bsen+%5Cbeta+%7D+%3D++%5Cfrac%7B82%7D%7B0%2C82%7D+---%5C+%5Ctextgreater+%5C+isolamos%5C+%28sen+%5Cbeta+%29%2C+fica.+%5C%5C++%5C%5C+sen+%5Cbeta+%3D+%5Cfrac%7B86%2C2.%280%2C82%29%7D%7B82%7D++%5C%5C++%5C%5C+sen+%5Cbeta+%3D0%2C862--%5C+%5Ctextgreater+%5C+por%5C+arcosen+%5C+de%5C++%5Cbeta+%2Ctemos.+%5C%5C++%5C%5C++%5Cbeta+%3D+sen%5E%7B-1%7D+0%2C862+%5C%5C++%5C%5C++%5Cbeta+%3D59%2C5418.....---%5C+%5Ctextgreater+%5C+aproximado+%5C%5C++%5C%5C+%5Cboxed%7B+%5Cboxed%7B%5Cbeta+%3D+59%2C5%5E%7Bo%7D+%7D%7D "\frac{x}{sen \beta } = \frac{82}{sen55^{o} } --\ \textgreater \ valor \ aproximado\ de[ sen55=0,82]\ e\ [ x=86,2] \\ \\ \frac{86,2}{sen \beta } = \frac{82}{0,82} ---\ \textgreater \ isolamos\ (sen \beta ), fica. \\ \\ sen \beta = \frac{86,2.(0,82)}{82} \\ \\ sen \beta =0,862--\ \textgreater \ por\ arcosen \ de\ \beta ,temos. \\ \\ \beta = sen^{-1} 0,862 \\ \\ \beta =59,5418.....---\ \textgreater \ aproximado \\ \\ \boxed{ \boxed{\beta = 59,5^{o} }}")
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Espero ter ajudado!!
Vamos calcular o (x) usando o lei senos , veja no anexo baixo
⇔⇔⇔⇔⇔⇔⇔⇔⇔⇔⇔⇔⇔⇔⇔⇔⇔⇔⇔⇔⇔⇔⇔
Já sabendo o valor de (x) calculamos o ângulo (β), resolvemos por lei de seno tambem, assim.
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Espero ter ajudado!!
Anexos:
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