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[tex]\boxed{\begin{array}{l}\sf o~exerc\cute icio~est\acute a~equivocado~pois\\\sf -1\leq sen\,x\leq1\\\sf logo~sen\,x=6~\acute e~um~absurdo.\\\sf contudo~vou~resolver~supondo~sen\,x=\dfrac{1}{6}.\\\sf como~\hat angulos~s\tilde ao~complementares\\\sf o~seno~de~um~\acute e~igual~ao~cosseno~de~outro.\\\sf portanto~sen(x)=cos(y)=\dfrac{1}{6}\\\sf vamos~calcular~sen~y: \\\sf cos\,y=\dfrac{1}{6}\implies cos^2\,y=\dfrac{1}{36}\\\sf sen^2\,y=\dfrac{36}{36}-\dfrac{1}{36}\\\sf sen^2\,y=\dfrac{35}{36}\\\sf \end{array}}[/tex]
[tex]\boxed{\begin{array}{l}\sf sen\,y=\dfrac{\sqrt{35}}{\sqrt{36}}\\\sf sen\,y=\dfrac{\sqrt{35}}{6}\\\sf tg\,y=\dfrac{sen\,y}{cos\,y}\\\sf tg\,y=\dfrac{\frac{\sqrt{35}}{6}}{\frac{1}{6}}\\\sf tg\,y=\diagup\!\!\!6\cdot\dfrac{\sqrt{35}}{\diagup\!\!\!6}\\\sf tg\,y=\sqrt{35}\end{array}}[/tex]
