## 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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1. anonymous

just wait im posting it

2. anonymous

3. anonymous

@Ashleyisakitty

4. anonymous

hmmm looks pretty straightforward what'st he value of "x"? what's the value of "y" in say 3a?

5. anonymous

x=- Square root of 3 and y= -1

6. jim_thompson5910

hint: make sure x^2+y^2 = 1

7. anonymous

o.

8. anonymous

ok.. so.. the radius or hypotenuse will be then $$\bf r=\sqrt{x^2+y^2}\implies r=\sqrt{(-\sqrt{3})^2+(-1)^2} \\ \quad \\ r=\sqrt{(-\sqrt{3})(-\sqrt{3})+1} \implies r=\sqrt{(\sqrt{3})^2+1} \\ \quad \\ r=\sqrt{3+1}\implies r=\sqrt{4} \implies r=2$$

9. jim_thompson5910

jdoe0001 has the most effective method, so go with that

10. anonymous

hmmm shoot

11. anonymous

$$\bf sin(\theta) =\cfrac{y}{r} \qquad % cosine cos(\theta) =\cfrac{x}{r} \qquad % tangent tan(\theta) =\cfrac{y}{x} \\ \quad \\ % cotangent cot(\theta) =\cfrac{x}{y} \qquad % cosecant csc(\theta) =\cfrac{r}{y} \qquad % secant sec(\theta) =\cfrac{r}{x}$$