Hard Math Problem

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
Science Advisor
Gold Member
7,161
54
Do you mean:

[tex] {(\sqrt 7)}^2 = r^2 + r^2 -2rr \cos{(\frac {2 \pi} 3)} [/tex]
?
 
622
2
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
Science Advisor
Gold Member
6,907
13
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 ?
 
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 ?
Nope! :biggrin:

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
 
111
0
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.
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top