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Mathematics 20 Online
OpenStudy (anonymous):

The point (-3, 1) is on the terminal side of angle ?, in standard position. What are the values of sine, cosine, and tangent of ?? Make sure to show all work.

OpenStudy (anonymous):

who ever answers this will get a medl fan and some owl buck !!!!!!!!!!!!!

OpenStudy (anonymous):

Let me see if i can do this...

OpenStudy (anonymous):

okay ! ;)

OpenStudy (anonymous):

You can make a right triangle using that point (and the line that leads to it, which is the terminal side) by going down to the x-axis and then over to the origin.

OpenStudy (anonymous):

continue

OpenStudy (anonymous):

You should now have a triangle with a base that is 3 units long and a height of 1 unit. If you use the Pythagorean Theorem, you can find the hypotenuse (usually this is just left as a radical if it doesn't come out to be a nice number).

OpenStudy (anonymous):

Do u understand so far?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

So can u find those sides?

OpenStudy (anonymous):

You should have a coordinate graph for this problem, or some graph paper...

OpenStudy (anonymous):

not really but iwas asking can you give me the answer for this and iwill medal fan and give you owlbuck = money ! but please this the last question then im done with alegebra 2

OpenStudy (anonymous):

Lol u want the answer. Ok, but give me a minute. It's a lot better if u understand it though

OpenStudy (anonymous):

It's 162!

OpenStudy (anonymous):

yea its but ineed to get this over with its been stressfullol

OpenStudy (anonymous):

I know how u feel btw...

OpenStudy (anonymous):

and dont we all butcan show like how you got 162 cause she needs work

OpenStudy (anonymous):

Work shown?

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

The length of the y component is √(1-0)^2 +(-3-(-3))^2] =√(12+ 02) = 1 A(-3,1) to B(-3,0) opposite The length of the x-component is √[(-3-0)^2 +(0-0)^2] = √(9+0)= 3 B(-3,0) to O(0,0) adjacent The length of vector OA is √[(-3-0)^2 + (1-0)^2] = √(9+1) = √(10) A(-3,1) to O(0,0) hypotenuse θ = 180 - α sinθ = sin(180-α) = opposite/hypotenuse = 1/√10 cosθ = adjacent/hypotenuse = -3/√10 tanθ = opposite/adjacent = 1/-3 = -1/3 α= arcsin(1/√10) ≈ 18 θ =180 -18 ≈162 WOW A LOT OF LETTERS!

OpenStudy (anonymous):

And numbers...

OpenStudy (anonymous):

@jammy987 u there?

OpenStudy (anonymous):

yess may god bless you bro !!!!!!!!!!!!

OpenStudy (anonymous):

The sin, cos, and tan values r right there... those r ur answers and ur welcome!

OpenStudy (anonymous):

okay ;)

OpenStudy (anonymous):

Np. :)

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