Polar coordinates of a point are given. Find the rectangular coordinates of the point. (-5, -180°) a. (-5,0) b. (0,-5) c. (0,5) d. (5,0)

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Polar coordinates of a point are given. Find the rectangular coordinates of the point. (-5, -180°) a. (-5,0) b. (0,-5) c. (0,5) d. (5,0)

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\(y=r\cdot \sin(\theta)\) \(x=r\cdot \cos(\theta)\)
You are given that: \(r=-5\) \(\theta = -180^\circ\)

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Other answers:

So, -5x-180?
Sorry, I am teaching myself. I am somewhat confused. I need help.
This is in general: \(y=r\cdot \sin(\theta)\) \(x=r\cdot \cos(\theta)\) (if you want to know why these values are what x and y are equivalent to, then let me know) And in our case, \(y=(-5)\cdot \sin(-180)=(-5)\cdot 0=0\) \(x=(-5)\cdot \cos(-180)=(-5)\cdot(-1)=5\)
Oh okay. I see now.
I am writing this down.
|dw:1444238217960:dw|
The length of the base is x, and the length of the side is y. This is (roughly) how cartesian coordinates are represented (ok?)
I am still here. I am writing this down.
And polar coordinates are represented differently: You face the direction of \(\huge _{^\theta}\) and you walk in this direction \(\huge _{^{_r}}\) units long.
If you have questions about anything that I am posting, please ask....
as it is now I will assume you got no questions.
I am understanding well.
And based on the diagram 9above), you can see that: \(\sin(\theta)={\rm opposite/hypotenuse}=y/r\) \(\cos(\theta)={\rm adjacent/hypotenuse}=x/r\) (this is simple trigonometry) Now, multiply both sides of each equation times r. You will get: \(\displaystyle \color{red}{ r \times}\sin(\theta)=\color{red}{ r \times}\frac{y}{r}\) \(\displaystyle\color{red}{ r \times}\cos(\theta)=\color{red}{ r \times}\frac{x}{r}\) (r cancels on the right side of each of the equations) \(\displaystyle r\sin(\theta)=y\) \(\displaystyle r\cos(\theta)=x\)
Thus, knowing the angle, you can convert any polar coordinate to cartesian coordinate (just as we did in the example ` (-5, -180°)` )
Okay, I see.
I mean: knowing the angle and the radius.
Are you doing calculus with polar coordinates, or this is some other course?
Pre-Calculus.
Oh, cool. You will encounter it more in calculus, but as it is right now, good luck!
And if anything there are many people who are willing to help with stuff.....
bye

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