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work out missing angle?
hi, i need to work out the angle in this shape:
Attachment 20114
im not really sure that to do but would this work:
cosA= 5^2 + 10^2 + 18^2/ 2x5x10= 449/100= 4.49?
is this correct? if not could you please tell me where im going wrong and what i need to do!?!
Thanks!

I don't think you're setting up your law of cosines correctly. Take a look at the wiki link, and tell me what the equation ought to be.

one question, where would A,B and C be on my triangle?

Take a really good look at the wiki link. It explains how to assign all the variables in the equation there. Then you tell me how to assign them.

would it be, ?^2= 7^2 + 5^22x7x5xcos38= 18.84, then square root it= 4.34?

Am I looking at the same picture you're looking at? I see a triangle with lengths 11, 18, and 10. You're asked to find the angle opposite the side of length 18. Is that correct?

yes, would it be:
18^2= 7^2+5^22x5x7xcos?= ??

Where is the 7 and 5 coming from in your equation?

GRRRR!!!! really sorry i was looking at the wrong question!
would it be:18^2= 10^2+11^22x10x11xcos?= ??

That looks better. Try solving for the cosine of the angle. What do you get?

im not really sure how i would do that.

Well, if you need to, just substitute x = cos(?) and solve for x the way you would normally do in algebra. What do you get?

i can't think im having a mental blank! grr!

could you tell me how to do it please, i can't think!

Yes, you can think. Your equation is this:
$\displaystyle 18^{2}= 10^{2}+11^{2}2\times 10\times 11\times \cos(?)$
Simplifying one step leads to
$\displaystyle 324=100+121220\cos(?)$
Your goal, as usual, is to get the $\displaystyle \cos(?)$ on one side of the equation and everything else on the other. What's your next step?