Let ABC be a triangle such that angle ACB = 135°. Prove that:
I really have no Idea how to solve this, any help is really appreciated. Thanks
That looks very similar to the law of cosines.. by the way you have to add spaces after \times when what follows are letters, otherwise it won't render. I fixed it while quoting you.
Let ABC be a triangle such that angle ACB = 135°. Prove that:
I really have no Idea how to solve this, any help is really appreciated. Thanks
Are you sure that's it? It is the cosine rule applied to the given triangle (or have I missed something?), in which case it looks a bit easy to be an IMO question
Let ABC be a triangle such that angle ACB = 135°. Prove that:
I really have no Idea how to solve this, any help is really appreciated. Thanks
You can also form a right-angled triangle by standing the triangle on side CB (or side AC).
Drop a vertical to O from A and join O to C such that OCB is a straight line.
This requires only Pythagoras' Theorem.
Then..
since
Finish with
Last edited by Archie Meade; July 20th 2010 at 02:33 PM.