Solving a Law of Cosines Problem: Finding the Length of a Side

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The discussion centers on solving a Law of Cosines problem to find the length of a side in a parallelogram. The initial attempt involved calculating a distance using the formula, resulting in a value of 3.23. Participants provided hints regarding the properties of parallelograms, specifically the relationships between opposite angles and the sum of angles. This guidance helped clarify the problem for the original poster, making it easier to approach. Understanding the basic properties of parallelograms was key to progressing in the solution.
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Homework Statement



[PLAIN]http://img28.imageshack.us/img28/1046/mathproblemp.jpg

Homework Equations



Law of Cosines

The Attempt at a Solution



I'm not really sure what to do for this problem, it looks easy but I can't figure it out. I tried # 11 and was able to find d.

d2=a2+b2-2(a)(b)cos(30)
d=3.23

However, I don't know where to go from there.
 
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Hints:

1. What do you know about the angles of the opposite corners of a parallelogram?
2. What do you know about the sum of the angles of a parallelogram?
 
Ahh... thanks, I didn't know the basic properties of parallelograms. Now this problem is much easier.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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