anonymous
  • anonymous
Two ocean liners leave from the same port in Puerto Rico at 10:00 a.m. One travels at a bearing of N 51o W at 14 miles per hour, and the other travels at a bearing of S 56o W at 17 miles per hour. Approximate the distance between them at noon the same day. Round answer to two decimal places.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
is it 51 degree north of west
anonymous
  • anonymous
?
anonymous
  • anonymous
A. 20.33 miles B. 43.25 miles C. 28.20 miles D. 22.72 miles E. 37.19 miles

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More answers

anonymous
  • anonymous
that doesnt seem reasonable......because it said round your answer to two decimal points. oh answers!
anonymous
  • anonymous
Find x and Y component of speed of each ship
anonymous
  • anonymous
ok so ship a would be 28 miles and shib b would be 34 miles i think
anonymous
  • anonymous
Here How I would solve it .. Ship 1 travel at 51 degree North of West Using standard coordinate system X component of speed is -cos(51)*14 Y component of speed is sin(51)*14
anonymous
  • anonymous
This is just ship 1
anonymous
  • anonymous
okay
anonymous
  • anonymous
try this with ship 2
anonymous
  • anonymous
so you will get speed of two ship
anonymous
  • anonymous
Multiply x and Y component of each ship seperately to find position then just use distance formula
anonymous
  • anonymous
ok so ship a is -97.49 ship b is 128.6 ?
anonymous
  • anonymous
X position of Ship 1 at noon(12-10am) -cos(51)*14 mile per hours * 2hour -17.62 Y position of Ship 2 at noon sin(51)*14*2=21.76
anonymous
  • anonymous
X position of Ship 2 -cos(56)*17*2=something Y position of Ship 2 -sin(56)*17*2=something Distance Formula \[\sqrt(ship 1 (x) position - ship 2( x) position)^2+ship 1 (y) position - ship 2( y) position)^2)\]
anonymous
  • anonymous
X position of Ship 2 -cos(56)*17*2=-23.89 Y position of Ship 2 -sin(56)*17*2=17.73
anonymous
  • anonymous
so then I just plug those into the formula?
anonymous
  • anonymous
I got 19.01 for x position of ship 2
anonymous
  • anonymous
hmmm ok you're probably right
anonymous
  • anonymous
Yep
anonymous
  • anonymous
can you see the distance formula in full
anonymous
  • anonymous
no
anonymous
  • anonymous
it is basically square root of (xposition of ship 1- xpos ship 2)^2 + (ypos ship 1-y pos ship 2)^2)
anonymous
  • anonymous
Is it physics or math?
anonymous
  • anonymous
math trig/pre calc
anonymous
  • anonymous
Were you able to get answer
anonymous
  • anonymous
?
anonymous
  • anonymous
ok I have to go back to the begining and replug in stuff cause i messed something up hold on
anonymous
  • anonymous
I will do the same thing
anonymous
  • anonymous
ok thanks
anonymous
  • anonymous
I got 45.73.....?
anonymous
  • anonymous
The answer is 37.19
anonymous
  • anonymous
I read that problem wrong it is 51 degree west of north ---- not north of west so I have to switch sin to cos vice versa
anonymous
  • anonymous
okay I have no clue what i did then
anonymous
  • anonymous
Ship 1 position X: - sin(51)*14*2 -- A Y: cos(51)*14*2 --B Ship 2 position X: -sin(56)*17*2 -- C Y: -cos(56)*17*2 --D
anonymous
  • anonymous
okay
anonymous
  • anonymous
sqrt((A-C)^2+(B-D)^2)
anonymous
  • anonymous
=37.19
anonymous
  • anonymous
ok thank you!

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