Determina la razón trigonométrica de la siguiente figura, para 60°

Explicación paso a paso:
[tex]h = \sqrt{ {8}^{2} - {4}^{2} } [/tex]
[tex]h = \sqrt{64 - 16} [/tex]
[tex]h = \sqrt{48} [/tex]
[tex]h = 4 \sqrt{3} [/tex]
[tex] \sin(60) = \frac{4 \sqrt{3} }{8} - - simplificar[/tex]
[tex] \sin(60) = \frac{ \sqrt{3} }{2} [/tex]
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[tex] \cos(60) = \frac{4}{8} - - - simplificar[/tex]
[tex] \cos(60) = \frac{1}{2} [/tex]
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[tex] \tan(60) = \frac{4 \sqrt{3} }{4} - - simplificar[/tex]
[tex] \tan(60) = \sqrt{3} [/tex]
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[tex] \csc(60) = \frac{2}{ \sqrt{3} }[/tex]
[tex] \csc(60) = \frac{2}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } [/tex]
[tex] \csc(60) = \frac{2 \sqrt{3} }{3} [/tex]
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[tex] \sec(60) = \frac{8}{4} [/tex]
[tex] \sec(60) = 2[/tex]
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[tex] \cot(60) = \frac{4}{4 \sqrt{3} } [/tex]
[tex] \cot(60) = \frac{1}{ \sqrt{3} } [/tex]
[tex] \cot(60) = \frac{ \sqrt{3} }{3} [/tex]
[tex]h^2=8^2-4^2\\\\h^2=64-16\\\\h^2=48\\\\h=\sqrt{48}\\\\h=\sqrt{2^4\cdot \:3}\\\\h=2^2\sqrt{3}\\\\h=4\sqrt{3}[/tex]
Determina la razón trigonométrica:
sec (60) ≈ 2
csc (60) ≈ [tex]\frac{2\sqrt{3} }{3}[/tex]
tan (60) ≈ √3
cot (60) ≈ [tex]\frac{\sqrt{3} }{3}[/tex]