SOLUTION: Two boats leave the harbour at the same time. Boat A travels at 32km/h on a bearing of 125 degrees and boat B travels at 18km/h on a bearing of 205 degrees After 3 hours How

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Question 543470: Two boats leave the harbour at the same time. Boat A travels at 32km/h on a bearing of 125 degrees and boat B travels at 18km/h on a bearing of 205 degrees
After 3 hours
How far apart are the boats?
In what direction would the skipper of boat B have to look to see boat A?
Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website!
Two boats leave the harbour at the same time. Boat A travels at 32km/h on a bearing of 125 degrees and boat B travels at 18km/h on a bearing of 205 degrees
After 3 hours
How far apart are the boats?
In what direction would the skipper of boat B have to look to see boat A?
**
Draw a triangle with the apex representing the starting point where the boats left the harbor. Call this point C. From this point C, extend a 96 km line at a bearing of 125�. Call this point A.
Also from point C, extend a 54 km line at a bearing of 205�. Call this point B. Draw a line connecting points A and B which is the distance(c) between boats A and B. You now have a triangle with two sides, CB and CA and their included angle of 80�. Solve with law of cosines:
..
Law of cosines: c^2=a^2+b^2-2*a*b cos80�
c^2=54^2+96^2-2*54*96 cos80�
c^2=2916+9216-10368 cos80�
c^2=12432-1800.38=10631.62
c=√10631.62≈103.11 km
..
sin B/96=sine C/103.11=sin 80�/103.11
sin B=(96/103.11)*sin80�=.917
B≈66.5�
Bearing of boat A from boat B=25+66.5=91.5�
..
ans:
The boats are 103.11 km apart after 3 hours.
The skipper of boat B must look to see boat A at a bearing of 91.5�
..
please check my calculations