A medida de um angulo agudo tal que sen ALFA tal que sen alfa = 12/13 calcule cos alfa e tg alfa
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Podemos utilizar a fórmula para descobrir o cos:
Substituindo:
![sen^{2}\alpha+cos^{2}\alpha = 1
\\\\
(\frac{12}{13})^{2}+cos^{2}\alpha =1
\\\\
\frac{144}{169}+cos^{2}\alpha = 1
\\\\
cos^{2}\alhpa = 1-\frac{144}{169}
\\\\
cos^{2}\alpha = \frac{169-144}{169}
\\\\\
cos^{2}\alpha = \frac{25}{169}
\\\\
cos\alpha = \pm \sqrt{\frac{25}{169}}
\\\\
cos\alpha = \pm \frac{5}{13}
\\\\
\boxed{angulo \ agudo < 90\°}
\\\\
\boxed{\boxed{cos\alhpa = +\frac{5}{13}}}](https://tex.z-dn.net/?f=sen%5E%7B2%7D%5Calpha%2Bcos%5E%7B2%7D%5Calpha+%3D+1%0A%5C%5C%5C%5C%0A%28%5Cfrac%7B12%7D%7B13%7D%29%5E%7B2%7D%2Bcos%5E%7B2%7D%5Calpha+%3D1%0A%5C%5C%5C%5C%0A%5Cfrac%7B144%7D%7B169%7D%2Bcos%5E%7B2%7D%5Calpha+%3D+1%0A%5C%5C%5C%5C%0Acos%5E%7B2%7D%5Calhpa+%3D+1-%5Cfrac%7B144%7D%7B169%7D%0A%5C%5C%5C%5C%0Acos%5E%7B2%7D%5Calpha+%3D+%5Cfrac%7B169-144%7D%7B169%7D%0A%5C%5C%5C%5C%5C%0Acos%5E%7B2%7D%5Calpha+%3D+%5Cfrac%7B25%7D%7B169%7D%0A%5C%5C%5C%5C%0Acos%5Calpha+%3D+%5Cpm+%5Csqrt%7B%5Cfrac%7B25%7D%7B169%7D%7D%0A%5C%5C%5C%5C%0Acos%5Calpha+%3D+%5Cpm+%5Cfrac%7B5%7D%7B13%7D%0A%5C%5C%5C%5C%0A%5Cboxed%7Bangulo+%5C+agudo++%26lt%3B+90%5C%C2%B0%7D%0A%5C%5C%5C%5C%0A%5Cboxed%7B%5Cboxed%7Bcos%5Calhpa+%3D+%2B%5Cfrac%7B5%7D%7B13%7D%7D%7D "sen^{2}\alpha+cos^{2}\alpha = 1
\\\\
(\frac{12}{13})^{2}+cos^{2}\alpha =1
\\\\
\frac{144}{169}+cos^{2}\alpha = 1
\\\\
cos^{2}\alhpa = 1-\frac{144}{169}
\\\\
cos^{2}\alpha = \frac{169-144}{169}
\\\\\
cos^{2}\alpha = \frac{25}{169}
\\\\
cos\alpha = \pm \sqrt{\frac{25}{169}}
\\\\
cos\alpha = \pm \frac{5}{13}
\\\\
\boxed{angulo \ agudo < 90\°}
\\\\
\boxed{\boxed{cos\alhpa = +\frac{5}{13}}}")
E tg é apenas sen sobre cos:
Substituindo:
E tg é apenas sen sobre cos:
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