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I have no idea how to do this problem can you guys shed some light?
Sqrt[7]^2 == r^2 + r^2 - 2 * r * r * Cos[2 * Pi / 3]
Thanks.
Sqrt[7]^2 == r^2 + r^2 - 2 * r * r * Cos[2 * Pi / 3]
Thanks.
Nope!Gokul43201 said:Is this the problem :
[tex]\sqrt{7}^2 = r^2 + r^2 - 2r^2 cos \left( \frac{2 \pi}{3} \right) [/tex] ?
Can you not simplify the LHS, collect terms in the RHS and plug in the value of the cosine to end up with a trivial quadratic ? Or did I misunderstand the question ?