Melody

What is wrtong with you MathCuber  ??

Your getting propfessional teaching help and you vote it down.  

Are you one of these cheating creeps who only is interested in the answer?

I thought you were better than that!

That is not finished and it is not what I got.

Can you show your working so I can see where you went wrong? 

             (mine could be wrong too but it worked out nicely so it probbly isn't)

Did you start it one side at a time or what?

You can find BC using cosine rule.  You have 2 sides and the included angle so you can find the 3rd side.

Your title is different from your question.

Please proof your questions properly and get rid of meaningless dollar signs.

Why don't you show me your answer?  I'd like that.

5+   5*5 = 30

You can take 9z of each of them.

Thanks MathCuber  

Attn: asker

this is a good concept for you to work out quickly.

\(\sqrt{169}=+13\qquad \text{(never -13, this is by convention)}\\ \)

However if 

\(y^2=169\)

then it is you that must introduce the square root sign,

     which means that both the negative and the positive are answer.

\(y^2=169\\ y=\pm\sqrt169\\ y=+13\;\;\;\;or\;\;\;\;y=-13\)

I may not have done this the best way but this was my method:

1)  Find BC

2)  find angle BCP

3) Find angle ACQ

4) Find angfle CAQ

5) Now you have 3 angles and one side of triangle ACQ so you can find AQ

This is a teaching answer. Please do not override it.

MathCuber, if you have questions/problems then please ask.

Hi MathCuber,

I did this without any trig until the very end.

Just expand and simplify for starters.

Then compare what you have to the cosine rule and you should be able to work it out.

If you have questions then ask.  Show or tell me what you have done though.

This is a teaching answer. Please noone interfer with a more full answer.