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kittiwitti1
 one year ago
A twoperson tent is to be made so that the height at the center is a = 6 feet (see the figure below). If the sides of the tent are to meet the ground at an angle 60°, and the tent is to be b = 8 feet in length, how many square feet of material will be needed to make the tent? (Assume that the tent has a floor and is closed at both ends, and give your answer in exact form.)
Picture reference:
http://prntscr.com/8a6azy
I did it two ways and got different answers, not sure which one I did wrong...
kittiwitti1
 one year ago
A twoperson tent is to be made so that the height at the center is a = 6 feet (see the figure below). If the sides of the tent are to meet the ground at an angle 60°, and the tent is to be b = 8 feet in length, how many square feet of material will be needed to make the tent? (Assume that the tent has a floor and is closed at both ends, and give your answer in exact form.) Picture reference: http://prntscr.com/8a6azy I did it two ways and got different answers, not sure which one I did wrong...

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440836698456:dw

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0I tried doing it on my own and got this:\[\frac{360}{\sqrt{3}}ft^{2}\]

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0I feel like that's wrong somehow...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, let's check.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You split the area into 3 separate sections, yes? The triangle faces and the rectangular sides?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And then added the base?

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0The base is the same as the sides in area, right? :o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440837218920:dw

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0These are the formulas I got. Triangle:\[h=6,b=\frac{6}{\sqrt{3}},hyp=\frac{12}{\sqrt{3}}\] Square:\[w=\frac{12}{3},l=8\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we need to find x first, and then double it to find the area of the front facing triangluar portion.

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0I think if I did anything wrong I probably messed up there

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I havent worked it out yet so one minute! :P

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440837311425:dw Okay lol

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440837440575:dw Do these look right to you?

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0I mean correctly done* lol that sounded rude. ^^: I found another proof online with different digits for A and B though, and got a different answer: http://mathhelpboards.com/questionsothersites52/aju051000squestionsyahooanswersinvolvingtrigonometry6263.html

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah. the \(\dfrac{12}{\sqrt{3}}\) looks correct, and that would be the side of the triangular prism... so \(A_2 = \dfrac{12}{\sqrt{3}} \cdot 8 \)

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0I got it like this:\[2\times\frac{1}{2}\times(6\times\frac{12}{\sqrt{3}})+3(8\times\frac{12}{\sqrt{3}})\]

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0Before the plus: Triangles. After the plus: Rectangles

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0I thought the link's method was too complicated and tried to do it myself...but the answers don't match. Not sure which one's wrong

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0I mean I got an answer of \[\frac{360}{\sqrt{3}}\] from doing it myself but it doesn't equal the answer I got from the method in the URL I posted:\[\ne\frac{240}{\sqrt{3}}\]

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0LOL Brb I have a mosquito in my house. _

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440838289662:dw \[SA = \color{red}{A_1} +\color{blue}{A_2} +\color{green}{A_3}\]\[SA = \color{red}{2\left(\frac{ 12}{\sqrt{3}} \cdot 8\right)} +\color{blue}{2\left(\frac{1}{2} \cdot \frac{12}{\sqrt{3}} \cdot 6\right)} + \color{green}{\left(\frac{12}{\sqrt{3}} \cdot 8\right)} \]

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0I used 3 instead of 2 for red because I just stuck all the rectangular areas together.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah that works too.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[SA = \color{red}{2\left(\frac{ 12}{\sqrt{3}} \cdot 8\right)} +\color{blue}{2\left(\frac{1}{2} \cdot \frac{12}{\sqrt{3}} \cdot 6\right)} + \color{green}{\left(\frac{12}{\sqrt{3}} \cdot 8\right)}\]\[SA = 3\left(\frac{12}{\sqrt{3}} \cdot 8 \right) +2\left(\frac{1}{2}\cdot \frac{12}{\sqrt{3}} \cdot 6\right)\]

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0\[3\left(8\times\frac{12}{\sqrt{3}}\right)+2\left(\frac{1}{2}\times6\times\frac{12}{\sqrt{3}}\right)=\frac{72}{\sqrt{3}}+3\left(\frac{96}{\sqrt{3}}\right)\]So far so good?

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0I mean is it good so far* (what is wrong with my grammar)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[SA = \frac{288}{\sqrt{3}} +\frac{72}{\sqrt{3}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then simplify it some more, and you get \[SA= \frac{360\sqrt{3}}{3} = 120\sqrt{3}\]

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0Yes. I did get that... But what's up with the other one giving me\[\frac{240}{\sqrt{3}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Just saw the link you posted earlier, the height being different would result in a different area for the triangle..hence different answer? that's why this guy is getting his \(SA = \dfrac{244}{\sqrt{3}}\).

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0a = 6ft b = 8ft θ = 60° \[\tan60°=\frac{a}{\frac{w}{2}}=\frac{2a}{w}\]\[w=2a\cot60°=\frac{8}{\sqrt{3}}ft\]\[\sin60°=\frac{a}{s}\]\[s=a\csc60°=\frac{8}{\sqrt{3}}ft\]

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0I did redo it with my own numerical values from my problem, but I still got a different answer.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OK let me work it out first, one min.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So they started ff by solving for half the base first, given the height of the triangle. dw:1440839311853:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry if im a ltitle slow in my responses right now, Kind of getting drowsy.

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0Okay, I'll type out the rest in case you guys can spot an error\[A_{T}=2\times\frac{1}{2}\times\frac{8}{\sqrt{3}}ft\times6ft=\frac{48}{\sqrt{3}}ft^{2}\]\[A_{R}=3\times\frac{8}{\sqrt{3}}\times8ft=64\sqrt{3}ft^{2}\] \[A=A_{T}+A_{R}=64\sqrt{3}ft^{2}+\frac{48}{\sqrt{3}}ft^{2}=\]\[64\sqrt{3}+\frac{48}{\sqrt{3}}\left(\frac{\sqrt{3}}{\sqrt{3}}\right)=\frac{48\sqrt{3}}{3}+64\sqrt{3}\]

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0BLEH that took too long...\[3\left(\frac{48\sqrt{3}}{3}\right)+3(64\sqrt{3})=48\sqrt{3}+192\sqrt{3}=240\sqrt{3}\]Not sure what I did wrong... And sorry it wasn't this \[\frac{240}{3}\]

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0*It wasn't 240/root 3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think however the guy solved it is a little too complicated and the way we solved it earlier is just easier. Don't you think?

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0So then for the next equation (which just has different values) I should do the same? http://prntscr.com/8a6nx4

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I think it was too longwinded and complex as well xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440839957574:dw \[\tan(60) = \frac{6}{\dfrac{w}{2}} = \frac{12}{w} \iff w = 12\cot(60^\circ)\] etc...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, now you just solve it with \(a=5\) instead of \(a=6\)

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, so I don't need to go the longwinded method... and just change the values in the equation right? :o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440840212338:dw and yes. I'll start you off here by helping you find the area of the triangular face.

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0It's okay, you're tired. I can handle this one :) I was afraid I had to do the long method but this one I can do. ^^

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh here's an easier way, if it helps at all. split the base of the triangle in half, and label it x. Then you have 2 x's. label the entire base y. So first you'll solve for x, and when you DOUBLE it, you'll find y :) Then you just use the yvalue in the rest of your calculations.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So what I would do is... dw:1440840514569:dw

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0I think I did that subconsciously xD

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0I got 6/sqrt 3 then x2 = 12/sqrt3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\tan(60) = \frac{5}{x} \iff x = \frac{5}{\tan(60)}\]\[y=2\left(\frac{5}{\tan(60)}\right)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0YEah, then you're good. no need for a long fancy explanation like that guy did.

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0What is the weird arrow you put there lol Is there a latex code?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That means its related. I could write either or interchangeably.

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0Oh. Is there a latex code x_x?dw:1440840784961:dw

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0Alright so far? (got the code thanks)

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0Thank you! :D I can handle it from here. Sorry to make you stay up to help me ^^;

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just remember that the triangle is equilateral, therefore the base will be the same as the hypotenuse.
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