Question

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

Answer 1

just wait im posting it

Answer 2

@Ashleyisakitty

Answer 3

hmmm looks pretty straightforward

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

Answer 4

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

Answer 5

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

Answer 6

o.

Answer 7

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

Answer 8

jdoe0001 has the most effective method, so go with that

Answer 9

hmmm shoot

Answer 10

\(\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}\)