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.
who ever answers this will get a medl fan and some owl buck !!!!!!!!!!!!!
Let me see if i can do this...
okay ! ;)
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.
continue
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).
Do u understand so far?
yes
So can u find those sides?
You should have a coordinate graph for this problem, or some graph paper...
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
Lol u want the answer. Ok, but give me a minute. It's a lot better if u understand it though
It's 162!
yea its but ineed to get this over with its been stressfullol
I know how u feel btw...
and dont we all butcan show like how you got 162 cause she needs work
Work shown?
yea
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!
And numbers...
@jammy987 u there?
yess may god bless you bro !!!!!!!!!!!!
The sin, cos, and tan values r right there... those r ur answers and ur welcome!
okay ;)
Np. :)
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