a) si los catetos son 2x y 2 x, hallamos su hipotenusa:
[tex] {(2x)}^{2} + {(2x)}^{2} = {h }^{2} \\ h = 2 \sqrt{2} x[/tex]
[tex]sen \alpha = \frac{2x}{2 \sqrt{2} x} = \frac{ \sqrt{2} }{2} [/tex]
[tex]cos \alpha = \frac{2x}{2 \sqrt{2} x} = \frac{ \sqrt{2} }{2} \\ \\ tan \alpha = \frac{2x}{2x} = 1[/tex]
b) sus catetos son √6 y √4
[tex] {( \sqrt{6)} }^{2} + {( \sqrt{4)} }^{2} = {h }^{2} \\ h = \sqrt{10} [/tex]
[tex] \sin(a) = \frac{ \sqrt{6} }{ \sqrt{10} } = \sqrt{ \frac{3}{5} } \\ \\ \cos(a) = \frac{ \sqrt{4} }{ \sqrt{10} } = \sqrt{ \frac{2}{5} } \\ \tan(a) = \frac{ \sqrt{6} }{ \sqrt{4} } = \sqrt{ \frac{3}{2} } [/tex]
el otro no veo :/
