anonymous
  • anonymous
I need help with #3. I can't figure out what im suppose to do with it.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
just wait im posting it
anonymous
  • anonymous
anonymous
  • anonymous
@Ashleyisakitty

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

jdoe0001
  • jdoe0001
hmmm looks pretty straightforward what'st he value of "x"? what's the value of "y" in say 3a?
anonymous
  • anonymous
x=- Square root of 3 and y= -1
jim_thompson5910
  • jim_thompson5910
hint: make sure `x^2+y^2 = 1`
anonymous
  • anonymous
o.
jdoe0001
  • jdoe0001
ok.. so.. the radius or hypotenuse will be then \(\bf r=\sqrt{x^2+y^2}\implies r=\sqrt{(-\sqrt{3})^2+(-1)^2} \\ \quad \\ r=\sqrt{(-\sqrt{3})(-\sqrt{3})+1} \implies r=\sqrt{(\sqrt{3})^2+1} \\ \quad \\ r=\sqrt{3+1}\implies r=\sqrt{4} \implies r=2\)
jim_thompson5910
  • jim_thompson5910
jdoe0001 has the most effective method, so go with that
jdoe0001
  • jdoe0001
hmmm shoot
jdoe0001
  • jdoe0001
\(\bf sin(\theta) =\cfrac{y}{r} \qquad % cosine cos(\theta) =\cfrac{x}{r} \qquad % tangent tan(\theta) =\cfrac{y}{x} \\ \quad \\ % cotangent cot(\theta) =\cfrac{x}{y} \qquad % cosecant csc(\theta) =\cfrac{r}{y} \qquad % secant sec(\theta) =\cfrac{r}{x}\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.