Question

Em um triângulo retângulo, a tangente de um de seus ângulos agudos
a é 7. Sabendo-se que a hipotenusa desse triângulo é 25, o valor do
seno desse mesmo ângulo é:

Answer 1

$\boxed{\begin{array}{l}\underline{\sf Observe\,a\,figura\,anexada.}\\\sf o\,enunciado\,nos\,diz\,que\,a\,tangente\\\sf de\,um\,dos\,\hat angulos\,agudos\,\acute e\,\dfrac{1}{7}\\\sf sendo\,assim\,irei\,considerar\, a\,tangente\,do\,\hat angulo\,\alpha.\end{array}}$

$\boxed{\begin{array}{l}\sf tg(\alpha)=7\\\sf mas~tg(\alpha)=\dfrac{x}{y}\\\sf7=\dfrac{x}{y}\longrightarrow y=\dfrac{x}{7}\\\sf pelo\,teorema\,de\,pit\acute agoras\,temos\\\sf x^2+y^2=25^2\\\sf x^2+(\dfrac{x}{7})^2=625\\\sf x^2+\dfrac{x^2}{49}=625\cdot49\\\sf 49x^2+x^2=30625\\\sf 50x^2=30625\\\sf x^2=\dfrac{30625}{50}\\\\\sf x=\dfrac{\sqrt{30625}}{\sqrt{50}}=\dfrac{\diagdown\!\!\!\!\!\!25\cdot7}{\diagdown\!\!\!\!5\sqrt{2}}=\dfrac{35}{\sqrt{2}}\\\\\sf x=\dfrac{35\sqrt{2}}{2}\end{array}}$

$\boxed{\begin{array}{l}\sf note\,que\,pelo\,tri\hat angulo\\\sf sen(\alpha)=\dfrac{x}{25}\\\\\sf sen(\alpha)=\dfrac{\frac{35\sqrt{2}}{2}}{25}=\dfrac{\diagup\!\!\!\!\!35^7\sqrt{2}}{2}\cdot\dfrac{1}{\diagdown\!\!\!\!\!\!25_5}\\\\\huge\boxed{\boxed{\boxed{\boxed{\sf sen(\alpha)=\dfrac{7\sqrt{2}}{10}}}}}\end{array}}$

Answer 2

$\large\displaystyle \sf \sf tg~ \alpha = \sf \dfrac{CO}{CA} = \dfrac {a}{b}$

$\large \displaystyle \sf 7 = \dfrac{a}{b} \Rightarrow \boxed{\displaystyle \sf a= 7b}$

$\large\boxed{\begin{array}{l} \displaystyle \sf c^{2}=a^{2} + b^{2}\\\displaystyle \sf 25^{2} = (7b)^{2} +b^{2}\\ \displaystyle \sf 25^{2} = 49b^{2} + b^{2}\\\displaystyle \sf 25^{2}= 50b^{2}\\ \displaystyle \sf b^{2}= \dfrac{25^{2} }{50} = \dfrac{2 \cdot25^{2} }{100} \\ \displaystyle \sf b = \dfrac{25\sqrt{2} }{10}=\dfrac{5\sqrt{2} }{2}\\ \displaystyle \sf a = 7b\\\displaystyle \sf a=\dfrac{7 \cdot 5 \cdot \sqrt{2} }{2 } \sf\end{array}}$

$\large\boxed{\begin{array}{l} \displaystyle \sf sen (a)= \dfrac{co}{hip} = \dfrac{a}{25} = \dfrac{ \dfrac{7 \cdot 5 \sqrt{2} }{2} }{ 25}\\\displaystyle \sf sen (a) =\dfrac{7 \cdot \sqrt{2} }{5 \cdot 2}\\\boxed{\displaystyle \sf sen (a) = \dfrac{7 \cdot \sqrt{2} }{10} } \sf\end{array}}$

Observe a figura anexada.

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