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AcidRa1n
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AcidRa1n
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Ok how does angle C equal 20 and PS is 5?
AcidRa1n
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Unkle it's not that hard come on...
UnkleRhaukus
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there are 360° in a complete revolution, angle C is 1 eighteenth of the wayround
\[\angle C=360°/18\]
AcidRa1n
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wait what so how its it 20 on there?
UnkleRhaukus
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did you simplify \(\angle C =\frac{360°}{18}=\dots\)
AcidRa1n
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ohhh
UnkleRhaukus
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\[36=2\times18\]
\[360=2 \times 18\times10\]
AcidRa1n
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How would you find the area?
UnkleRhaukus
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well the shape can be see as 18 of those little right angled triangle joined together,
find the area of one then multiply by 18
UnkleRhaukus
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|dw:1338432746562:dw|
AcidRa1n
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ok wait
AcidRa1n
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I have a diferent problem on my notes.. Same shape but it says each side = 4
AcidRa1n
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What is the area.
AcidRa1n
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|dw:1338433164896:dw|
AcidRa1n
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like that
UnkleRhaukus
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you can find length CS = a
using a trigonometric function of the angle 20°
AcidRa1n
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Lets say the question only give you the length of one side is 4
AcidRa1n
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How would you go about finding the area?
UnkleRhaukus
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|dw:1338432905847:dw|
AcidRa1n
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|dw:1338433361442:dw|
AcidRa1n
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Thats all it gives you. Find the area. I know it has something to do with finding the apothem and finding the perimeter and using the TAN function.
AcidRa1n
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Half of one side is 2.
UnkleRhaukus
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\[\tan(20°)=\frac{\text{opposite}}{\text{adjacent}}=\frac5a\]
\[a=\frac 5{\tan(20°)}\neq4\]
UnkleRhaukus
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or is the 4 coming from a different problem/
UnkleRhaukus
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the diagram posed at the top of the page does not have any side-
length equal to 4
AcidRa1n
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4 from a different problem
UnkleRhaukus
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OK, what is the new problem exactly/
AcidRa1n
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What I drew that's all it gives me
AcidRa1n
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A decagon with a side lengths of 4. Find the Area.
UnkleRhaukus
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well the first one was a nonagon so the picture is wrong
AcidRa1n
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RAWRRRRRRRRR I HATE MATH SO MUCH OMFG
UnkleRhaukus
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|dw:1338433609250:dw|
AcidRa1n
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I DONT HAVE TIME FOR THISSSS I WANNA SLEEP
AcidRa1n
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Yes now find the area. Just with that information.
UnkleRhaukus
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well should i sing you a lullaby instead ?
AcidRa1n
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LOL
UnkleRhaukus
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ok so you might want to find the angle in triangle first,
remember there are 360° in a revolution and i count 10 angles,
so the angle in is \[360°/10\]
UnkleRhaukus
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next find the length of the perpendicular , using the angle you just found and the tangent function
UnkleRhaukus
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|dw:1338433908040:dw|, find the area of the triangle and lastly multiply this area by the number of those triangles in the decagon
AcidRa1n
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360/10 =36? When I apply the TAN to 36 it's weird...
AcidRa1n
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@UnkleRhaukus
UnkleRhaukus
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true, what did you determine as an approximation of length b
UnkleRhaukus
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2 significant figures is probably fine
AcidRa1n
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huh? I didnt get that far becuase I thought you need to know the TAN of 36
AcidRa1n
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@satellite73
UnkleRhaukus
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yeah \[\tan(36°)\approx0.727\]
AcidRa1n
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ok.. Let me see
UnkleRhaukus
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\[\tan(36°)=\frac{\text{opp}}{\text{adj}}=\frac 2 b\]
\[b=\frac {2}{\tan36°}\approx\cdots\]
AcidRa1n
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why is it b= 2/tan36?
AcidRa1n
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Ohh I got it
AcidRa1n
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so its 2.75
UnkleRhaukus
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we are trying to find \(b\) so we can work out the are of the triangle \[\tan(36°)=\frac{\text{opp}}{\text{adj}}=\frac 2 b\]
multiply both sides by \(b\)
then divide both sides by \(\tan36°\)
UnkleRhaukus
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yes \(b\approx 2.75\)
AcidRa1n
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so then after we find the area which is 2.752 we take the perimeter which is 40, and do 1/2* 2.752* 40
UnkleRhaukus
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find the area of this triangle |dw:1338435087210:dw|
UnkleRhaukus
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\[A_{\triangle}=\frac{ab}2=\frac {2b}2=\cdots\]
UnkleRhaukus
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you dont need the perimeter
AcidRa1n
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My notes has it O.o
AcidRa1n
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Maybe hes teaching it like that?
AcidRa1n
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THe formula for Area is. 1/2 * Apothem * perimeter
UnkleRhaukus
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whatever, my method is better ~
so the area of the small right angled triangle is 2.752
and there are 20 of these in the decagon
the the total area is simply \(A_{decagon}=2.752\times20=\cdots\)
AcidRa1n
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Yeah lol same thing I got.
AcidRa1n
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Well, thanks for taking like 30 minutes of your time to help me lol.
AcidRa1n
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I'm so dumb.