# Hard Math Problem

#### IndustriaL

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.

#### Integral

Staff Emeritus
Gold Member
Do you mean:

$${(\sqrt 7)}^2 = r^2 + r^2 -2rr \cos{(\frac {2 \pi} 3)}$$
?

#### Gale

well first of all, thats the law of cosines. if i were you, i'd draw your triangle out, and use the law of sines which is simpler.
if you really need to solve it that way, its pretty simple. multiply everything out, and then factor out r^2 and solve.

not really sure where you got stuck, but you also have to show your work here.

#### Gokul43201

Staff Emeritus
Gold Member
Is this the problem :

$$\sqrt{7}^2 = r^2 + r^2 - 2r^2 cos \left( \frac{2 \pi}{3} \right)$$ ?

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 ?

#### TenaliRaman

Gokul43201 said:
Is this the problem :

$$\sqrt{7}^2 = r^2 + r^2 - 2r^2 cos \left( \frac{2 \pi}{3} \right)$$ ?

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 ?
Nope!

Also, 1-cos(theta) = 2sin^2(theta/2)
which simplifies most of the things for you, infact i think you can do it with your hands tied behind your back.

-- AI

#### wisredz

What is asked actually? to find r? Or something related to the triangle? If it is to find r, cos(2pi/3)=-1/2. the rest is easy.

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