Solve a Triangle using Trigonometry

[okh]

Homework Statement

http://img23.imageshack.us/img23/9427/imageyakj.jpg
AC+CB=24
The bisector = 8.
Find angle DCB, i.e. x.

Homework Equations

The Law of sines, I think.

The Attempt at a Solution

 

[praharmitra]

You are right by guessing Law of sines. Why don't you show what work you've done till now and where exactly you're stuck?

 

[okh]

I've written this system of equations:
http://img13.imageshack.us/img13/9355/imagenxvc.jpg

But I get
3cos^2x+sin^2x-2cosx=0 and the solution of the poblem that the book gives me is ∏/5.

 

[praharmitra]

I've written this system of equations:
http://img13.imageshack.us/img13/9355/imagenxvc.jpg

But I get
3cos^2x+sin^2x-2cosx=0 and the solution of the poblem that the book gives me is ∏/5.

The first three equations you've written down are correct. The answer is indeed pi/5. Check your calculation again. If you still don't find your mistake, write down your step by step solution, and i'll tell you where you're going wrong.

 

[okh]

The first three equations you've written down are correct. The answer is indeed pi/5. Check your calculation again. If you still don't find your mistake, write down your step by step solution, and i'll tell you where you're going wrong.

I've found my mistake, now I get 3cos^2x-sin^2x-2cosx=0 and the solution is pi/5.
Thank you very much for helping!

 

[Curious3141]

A word of advice, though. It's never a good idea to mix two different angle measures. Since you started in degrees, you should finish in the same. So the answer is x = 36 degrees.

 

[DryRun]

I agree, with Curious3141. You might also want to justify why you rejected the other value of x, which is 108 degrees.

I see 2 reasons:
1. Since x is an acute angle from the figure, x cannot be 108 degrees.
2. If you evaluate the angle CAD, using x=108 degrees, you'll get a negative value, which is not possible.

I hope that helped you a little bit.