I need help with #3. I can't figure out what im suppose to do with it.

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I need help with #3. I can't figure out what im suppose to do with it.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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just wait im posting it

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hmmm looks pretty straightforward what'st he value of "x"? what's the value of "y" in say 3a?
x=- Square root of 3 and y= -1
hint: make sure `x^2+y^2 = 1`
o.
ok.. so.. the radius or hypotenuse will be then \(\bf r=\sqrt{x^2+y^2}\implies r=\sqrt{(-\sqrt{3})^2+(-1)^2} \\ \quad \\ r=\sqrt{(-\sqrt{3})(-\sqrt{3})+1} \implies r=\sqrt{(\sqrt{3})^2+1} \\ \quad \\ r=\sqrt{3+1}\implies r=\sqrt{4} \implies r=2\)
jdoe0001 has the most effective method, so go with that
hmmm shoot
\(\bf sin(\theta) =\cfrac{y}{r} \qquad % cosine cos(\theta) =\cfrac{x}{r} \qquad % tangent tan(\theta) =\cfrac{y}{x} \\ \quad \\ % cotangent cot(\theta) =\cfrac{x}{y} \qquad % cosecant csc(\theta) =\cfrac{r}{y} \qquad % secant sec(\theta) =\cfrac{r}{x}\)

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