Calc
posted by Bill .
At noon, ship A is 100 Kilometers due east of ship B, Ship A is sailing west at 12 k/h, and ship B is sailing S10degrees west at 10k/h. At what time will the ships be nearest each other and what time will the distance be?
(not a right triangle)?

Not a right triangle?
You know angle, or SAS.
you want the other distance (S)
I think I would start with the law of cosines:
c^2=a^2+b^2 2ab CosC 
Ok. So c^2=100^2+b^22(100)b Cos100 ?

c^2=100^2x+(12t)^22(100x)(12t) Cos100 or maybe this?

Still trying to figure this out.
Am i going in the right direction?
c2=(12t)^2+(10t)^22(12t)(10t)Cos100
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