Resposta :
a)[tex]sen70 = \frac{cateto-oposto}{medida-da-hipotenusa} \\\\
0,94 = \frac{8}{x} \\\\
0,94x = 8\\\\
x = \frac{8}{0,94} \\\\
x = aproximadamente = 8,51[/tex]
b)[tex]cos65 = \frac{cateto-adjacente}{medida-da-hipotenusa} \\\\ 0,42 = \frac{x}{53} \\\\ x = 0,42.53\\\\ x = 22,26[/tex]
c)[tex] a^{2} = b^{2} + c^{2} \\\\ x^{2} = 6^{2} + 3^{2} \\\\ x^{2} = 36 + 9\\\\ x^{2} = 45\\\\ x = \sqrt{45} \\\\ x= 3 \sqrt{5} [/tex].
b)[tex]cos65 = \frac{cateto-adjacente}{medida-da-hipotenusa} \\\\ 0,42 = \frac{x}{53} \\\\ x = 0,42.53\\\\ x = 22,26[/tex]
c)[tex] a^{2} = b^{2} + c^{2} \\\\ x^{2} = 6^{2} + 3^{2} \\\\ x^{2} = 36 + 9\\\\ x^{2} = 45\\\\ x = \sqrt{45} \\\\ x= 3 \sqrt{5} [/tex].