cutiecomittee123
  • cutiecomittee123
use the law of cosines to find the value of 2-2-2 cos theta.
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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cutiecomittee123
  • cutiecomittee123
|dw:1449796541064:dw|
cutiecomittee123
  • cutiecomittee123
I am familar with the law of cosines formulas standard a^2=b^2-c^2bcos(A) b^2=a^2+c^2-2acos(B) c^2=a^+b^2-2abcos(C) alternat form Cos(A)=(b^2+c^2-a^2)/(2bc) Cos(B)=(a^2+c^2-b^2)/(2ac) Cos(C)=(a^+b^2-c^2)/(2ab) but the problem is, I am not sure which one to apply to this question.
cutiecomittee123
  • cutiecomittee123
btw pictured is an isoclese triangle

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cutiecomittee123
  • cutiecomittee123
@johnweldon1993
IrishBoy123
  • IrishBoy123
|dw:1449797665631:dw| you can apply it a lot of ways here you want \(\theta\)
cutiecomittee123
  • cutiecomittee123
yeh that is mostly what i needed is labeling as to which side is a, b, and c on the trianlge. Now i am confused on which law of cosines equation to use
IrishBoy123
  • IrishBoy123
which is A in the easy to remember version of the Law so \(a^2 = b^2 + c^2 - 2bc \cos A\) \(\cos A = \dfrac{b^2 +c^2 - a^2}{2bc}\) and \(b = c\)
cutiecomittee123
  • cutiecomittee123
so you can use standard or alternative to solve this?
IrishBoy123
  • IrishBoy123
"standard or alternative" ??
IrishBoy123
  • IrishBoy123
don't know the jargon πŸ˜–
cutiecomittee123
  • cutiecomittee123
yeah the forms, there is standard and alternative forms for law of cosine. Well there ya go, now you know lol
cutiecomittee123
  • cutiecomittee123
i listed them earlier in the thread
IrishBoy123
  • IrishBoy123
no, i still don't know!!!! there is just the law of cosines πŸ˜‹
IrishBoy123
  • IrishBoy123
some letters and an idea!!
IrishBoy123
  • IrishBoy123
oh!! that's just algebraic manipulation...... sorry, i skipped over it
cutiecomittee123
  • cutiecomittee123
1 Attachment
cutiecomittee123
  • cutiecomittee123
okay so now I just plug in the 1, 2, 2 into the equation and use cos^1 right?
IrishBoy123
  • IrishBoy123
thank you for sharing that!! yeah, plug it all in and fingers crossed.
cutiecomittee123
  • cutiecomittee123
nope it didnt work out:(((
cutiecomittee123
  • cutiecomittee123
@IrishBoy123
IrishBoy123
  • IrishBoy123
your question is \(2-2-2 \qquad \cos \theta\) but your drawing is 2-2-1, i think https://gyazo.com/82e54f67f6bbc11ea82d360a6ff4a2c9 do we need to get the question right first?!?!
cutiecomittee123
  • cutiecomittee123
the question is right! I thought that was confusing too but that i what the origional question looks like, I even double checked
IrishBoy123
  • IrishBoy123
so it's an equilateral triangle: 2-2-2 ?!?!
cutiecomittee123
  • cutiecomittee123
1 Attachment
anonymous
  • anonymous
if it is an equilateral triangle then all the angles are equal otherwise \[\theta=\cos^{-1} \left( \frac{ 4+4-1 }{ 2*2*2 } \right)=\cos^{-1} \left( \frac{ 7 }{ 8 } \right)=?\]
IrishBoy123
  • IrishBoy123
it's 7 another strangely worded question...
anonymous
  • anonymous
\[\theta=29degree \approx.\]
IrishBoy123
  • IrishBoy123
\(1^2 = 2^2 + 2^ 2 - 2(2) (2) \cos \theta \) \(2(2) (2) \cos \theta = 2^2 + 2^ 2 - 1^2\) \(\color{red}{2*2*2 \cos \theta} = 2^2 + 2^ 2 - 1^2 \color{red}{= 7}\)
cutiecomittee123
  • cutiecomittee123
okay i see it now, thanks:)
anonymous
  • anonymous
you want 2*2*2 cos theta only and irish boy's solution is perfect i thought you wanted theta.
IrishBoy123
  • IrishBoy123
@surjithayer me too
IrishBoy123
  • IrishBoy123
thank you @cutiecomittee123 i'd, FWIW, recommend just getting a handle on the idea behind the law of cosines. you can always ask here, tooπŸ˜‡

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