Asked 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)?
Answers
Answered by
bobpursley
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
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
Answered by
Bill
Ok. So c^2=100^2+b^2-2(100)b Cos100 ?
Answered by
Bill
c^2=100^2-x+(-12t)^2-2(100-x)(-12t) Cos100 or maybe this?
Answered by
Bill
Still trying to figure this out.
Am i going in the right direction?
c2=(12t)^2+(10t)^2-2(12t)(10t)Cos100
Am i going in the right direction?
c2=(12t)^2+(10t)^2-2(12t)(10t)Cos100
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