Question 144643: Hi, if anyone can tell me if I am in the right direction with this problem please.
Two boats leave a marina at the same time. Boat A travels at 20km/hr in a direction of 65 degrees. Boat B travels at 12.5km/hr in a direction of 145 degrees. How far apart are the boats after 2 hours. (2) In what direction would the skipper of Boat A have to look to see Boat B. I have this.
Boat A 20km x 2 hrs = 40 km
Boat B 12.5 km x 2 hrs = 25 km.
They are offset by an angle of 145-60 = 86 degrees. Using the Law of Cosines I get
x^2 = 40^2 + 25^2 - 2(40)(25)cos(85 degrees) = 2050.688. so x^2 = 45.28. The boats are 45.28 km apart after 2 hours. However this just doesnt seem right to me.
I have no idea how to answer part b.
Thanks.
Found 3 solutions by ankor@dixie-net.com, stanbon, Alan3354:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Two boats leave a marina at the same time. Boat A travels at 20km/hr in a direction of 65 degrees. Boat B travels at 12.5km/hr in a direction of 145 degrees. How far apart are the boats after 2 hours. (2) In what direction would the skipper of Boat A have to look to see Boat B. I have this.
:
I think you have the right idea here:
:
Boat A 20km x 2 hrs = 40 km
Boat B 12.5 km x 2 hrs = 25 km.
However the angle should be
They are offset by an angle of 145-65 = 80 degrees.
:
Using the Law of Cosines you should get:
:
x^2 = 40^2 + 25^2 - 2(40)(25)cos(80 degrees) = 1877.70 so x = 43.33.
The boats are 43.33 km apart after 2 hours.
:
Draw a diagram of this and you will see that Skipper A would have to look south-west to see boat B
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! Two boats leave a marina at the same time. Boat A travels at 20km/hr in a direction of 65 degrees. Boat B travels at 12.5km/hr in a direction of 145 degrees. How far apart are the boats after 2 hours. (2) In what direction would the skipper of Boat A have to look to see Boat B. I have this.
Boat A 20km x 2 hrs = 40 km
Boat B 12.5 km x 2 hrs = 25 km.
They are offset by an angle of 145-65 = 80 degrees. Using the Law of Cosines I get
d^2 = 40^2 + 25^2 - 2(40)(25)cos(80 degrees) = 1877.70. so d = 43.33 km.
The boats are 43.33 km apart after 2 hours.
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(2) In what direction would the skipper of Boat A have to look to see Boat B.
Find the angle (theta) formed by the 43.33 and the 25 km sides:
cos(theta) = (25^2+43.33^2-40^2)/(2*25*43.33) = 0.416565...
theta = 65.3826.. degrees
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Assuming you have drawn a picture of the problem on an x/y coordinate
system, draw a line parallel to the x-axis thru the vertex of the point
where boat B is. The 25 km side is a trasversal between the two parallel
lines.
The total angle at B is 65.3826 degrees; the alternate interior angles
formed by the transversal are 180-145 = 35 degrees.
So the line of sight from point B to point A is 65.38-35 = 30.38 degrees.
The line of sight from point A to B is 180 + 30.38 = 210.38 degrees
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Cheers,
Stan H.
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Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! You did it right, but the angle between the paths is 80 degs, not 86 or 85. That makes the distance 43.3325 km.
Then, use the law of sines to determine the angle from boat A to B, which is 34.623 degs. Looking from boat A back to the marina would be the recip or 65 degs, or 245 degs. Subtract the 34.623 from that, giving 211.377 degs.
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