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ksaimouliBest ResponseYou've already chosen the best response.0
dw:1358806909531:dw
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.0
what are you trying to find
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
dw:1358880151258:dwYour picture is so sloppy :D Lemme see if this looks correct.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Theta is the angle. s is the letter we use for arc length.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Umm let's first find the radius. I think we're going to need that. Here is a helpful formula:\[\large s=r \theta\]We want to solve for r so let's divide both sides by \(\theta\).\[\large \frac{s}{\theta}=r\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Our theta is given in degrees, so we'll need to convert it to radians.\[\large 30^o \left(\frac{\pi}{180^o}\right)=\frac{\pi}{6}\] So our radius is given by this,\[\large r=\frac{6\pi}{\left(\dfrac{\pi}{6}\right)}\] Which tells us that we have a radius of 36.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
So now we need the area of this sector. Hmm. So it's a percentage of the area of a circle. So it will end up being the same formula we use for area of a circle, just with a fraction in front of it to show that we only want a fraction of that area.\[\large A=\left(C\right)\pi r^2\]From here we just need to find out what C is. You actually probably have a formula on a sheet somewhere, so I likely don't need to go into this much detail :) lol my bad. Anyway, our fraction will be the number of degrees we care about DIVIDED BY the total degrees in a circle.\[\large A=\left(\frac{30^o}{180^o}\right)\pi r^2\]The degree bubbles will cancel out, giving us a unitless quantity (which is what we want).\[\large A=\left(\frac{1}{6}\right)\pi r^2\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
And we determined earlier that our radius was \(\large r=36\). Plugging this in gives us,\[\large A=\left(\frac{1}{6}\right)\pi (36)^2 \qquad = \qquad \left(\frac{1}{\cancel{6}}\right)\pi (6)^{\cancel{4}3}\]
 one year ago
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