# Hard Math Problem

1. Jun 13, 2005

### 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.

2. Jun 13, 2005

### Integral

Staff Emeritus
Do you mean:

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

3. Jun 13, 2005

### 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.

4. Jun 13, 2005

### Gokul43201

Staff Emeritus
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 ?

5. Jun 13, 2005

### TenaliRaman

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

6. Jun 14, 2005

### 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.