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sin(θ) =
cos(θ) =
csc(θ) =
sec(θ) =
cot(θ) =

Does that seem right so far?

I can keep going XD

So x^2 + y^2 = r^2 (r is the hypotenuse)
(1)^2 + (sqrt15)^2 = r^2
1+ 15=r^2
16=r^2
sqrt16=r
r=4

So now you have all the needed numbers to finish the problem.
x = 1
y = sqrt15
r = 4

sin = sqrt15/4
cos = 1/4
tan = sqrt15
csc = 4/sqrt15
sec = 4
cot = 1/sqrt15

Thank you so much! But the Sin= sqrt15/4 is not the right answer.

I can't seem to figure that one out.

Is everything else right?

Yes

Is it (sqrt15)/4 or is the 4 inside the sqrt of 15?

It is (sqrt15)/4 because it is y/r.

Ok, yeah that's the right answer. Thanks again :)

Oh okay, welcome! XD

you got it?

ok...gonna need the picture tho :)

A is the Adjacent angle, y is the hypotenuse, and x is the opposite. Acute triangle.

Yes, that pic.

LIKE THIS? opps caps were on lol

The A is on the bottom, but everything else is in the right place.

are we dealing with a right triangle?

yes

A is bottom left angle?

Just the bottom angle. I'll try to put up the pic

ahhh...thats better, when you said A was the adjacent angle it threw me :)

Sorry :[

'sok....ill live ;)

x is equal to "sin(t)"
sin(t) = x/y
x = y * sin(t)

probably want my to use A for that eh....

A = 4...that helps....

I don't understand how to use it in the equation.

you know how the trig functions are defined?

Not really.

What do I do for this problem though?
\

you just express x or y in terms of the trig functions... as far as I can tell

...simply the *ratio* of the .....

oonce you know sin and cos, the rest are just made with these two...

How would I find x and y then?

That's all it shows in the problem.

no angle or anything.

Here is another similar one if you can help with this one.
Find the side labeled x. Assume a = 4.

How would I express them in terms then?

this we have enough information to solve :)

well, I wrote up all the ones I could think of for the x values up there.

Oh.. ok I get this one. But the other has me confused :P

It just says "Express x in terms of trigonometric ratios of θ."

Same for y!

How would I go about doing that? I tried the X ones you put up, but they don't seem to work.

do you have anymore options :)

nope :(

then I would just skip it and move to the next question..... :)

lol .... well, I do have the time.....alot of time :)

Are you in college or what?