# Thread: work out missing angle?

1. ## work out missing angle?

hi, i need to work out the angle in this shape:

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!

2. 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.

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

4. 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.

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

6. 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?

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

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

9. GRRRR!!!! really sorry i was looking at the wrong question!

would it be:18^2= 10^2+11^2-2x10x11xcos?= ??

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

11. im not really sure how i would do that.

12. 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?

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

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

15. Yes, you can think. Your equation is this:

$18^{2}= 10^{2}+11^{2}-2\times 10\times 11\times \cos(?)$

$324=100+121-220\cos(?)$
Your goal, as usual, is to get the $\cos(?)$ on one side of the equation and everything else on the other. What's your next step?