Question 418268
Two ships leave the same port at 10:00 A.M. Ship A travels at the constant rate of 25 MPH at a bearing of S70E; Ship B travels at a constant rate of 20 MPH at a bearing of N29W. How far apart are the ships at noon?
b. what is the bearing to Ship B from Ship A at noon?

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At noon ship A would have traveled 2*25=50 miles.
At noon ship B would have traveled 2*20=40 miles.

Draw an (x,y) coordinate graph with center where both ships left at 10 AM from 
the same port.Ship A with a bearing of S70E makes a 20 degree angle with the x-axis. Ship B with a bearing of N29W makes a 29 degree angle with the y-axis.
You now have a triangle with sides of 50 mi and 40 mi with their included angle = 20+90+29=139 degrees.
Using Law of Cosines: c^2=a^2+b^2-2abCosx=50^2+40^2-2*50*40*cos 139 deg.
c^2=4100-(-3018.34)=7118.34
c=84.37 miles
ans:
The ships are 84.37 miles apart at noon.

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bearing to ship B from ship A
sinx/sin40=sin139/84.37
sinx=40sin139/84.37=.311
x=18.12 degrees
bearing angle=90-20-18.12=51.88 deg
ans:
After 12 noon,bearing to ship B from ship A=N51.88W