Question

Suppose that a triangle ABC has sides a,b,c. Prove the cosine rule, that is: a^2= b^2 + c^2 -2bc(cos(θ))

Answer 1

Can I use vectors?

Answer 2

Yes please.

Answer 3

Yes pretty please with cherries on top my master.

Answer 4

Take a vector triangle a+b=c and square

c.c = a.a + b.b +a.b +b.a ->

|c^2| = |a|^2 + |b|^2 + 2 a.b

Now convert to scalar form with a = |a|....and a.b = -ab Cos theta ->

c^2 = a^2 + b^2 - 2ab Cos C

Answer 5

U might want to tart it up a bit for a proof...

eg a.b =b.a and a.a = |a|^2 >= 0

Answer 6

One other thing, why does a.b = -abcosθ?

Answer 7

http://en.wikipedia.org/wiki/Dot_product

Answer 8

okay thanks alot.

Answer 9

ur welcome