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septicnation2
well sin cos and tan can only work on a right trangle so whats 90 - 58 = ?

its 32

dalendrk
[tex]\cos(90^o-\theta)=\sin\theta\ and\ \sin\theta=\cos58^o\\\\therefore\\\\90^o-\theta=58^o\ \ \ \ |subtract\ 90^o\ from\ both\ sides\\-\theta=-32^o\ \ \ |change\ signs\\\boxed{\theta=32^o}[/tex]

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