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?

Question

(not a right triangle)?

Answer 1

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

Answer 2

Ok. So c^2=100^2+b^2-2(100)b Cos100 ?

Answer 3

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

Answer 4

Still trying to figure this out.

Am i going in the right direction?

c2=(12t)^2+(10t)^2-2(12t)(10t)Cos100