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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just wait im posting it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm looks pretty straightforward what'st he value of "x"? what's the value of "y" in say 3a?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x= Square root of 3 and y= 1

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0hint: make sure `x^2+y^2 = 1`

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok.. so.. the radius or hypotenuse will be then \(\bf r=\sqrt{x^2+y^2}\implies r=\sqrt{(\sqrt{3})^2+(1)^2} \\ \quad \\ r=\sqrt{(\sqrt{3})(\sqrt{3})+1} \implies r=\sqrt{(\sqrt{3})^2+1} \\ \quad \\ r=\sqrt{3+1}\implies r=\sqrt{4} \implies r=2\)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0jdoe0001 has the most effective method, so go with that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf sin(\theta) =\cfrac{y}{r} \qquad % cosine cos(\theta) =\cfrac{x}{r} \qquad % tangent tan(\theta) =\cfrac{y}{x} \\ \quad \\ % cotangent cot(\theta) =\cfrac{x}{y} \qquad % cosecant csc(\theta) =\cfrac{r}{y} \qquad % secant sec(\theta) =\cfrac{r}{x}\)
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