Calculating Net Displacement Vector for Sparky's Run Using Law of Cosines

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To calculate Sparky's net displacement vector, the first step is to break down his movements into components. Sparky runs 50.0m northeast, which translates to approximately 35.4m east and 35.4m north. He then moves 70.0m west and 20.0m south, which should be subtracted from his initial position. Using these components, the final position can be determined, and the resultant displacement vector can be calculated. Redrawing the diagram to scale can also aid in visualizing the problem more effectively.
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Jim's dog Sparky runs 50.0m northeast to a tree, then 70.0m west to a second tree, and finally 20.0m south to a third tree.Calculate Sparky's net displacement vector.
I drew a picture and got an unknown triangle, which yileds the use of law of cos. I applies law of cos, to find an angle, and i get an angle to be 180 degrees. I know this value is wrong, but i know I will need at least one measure of an angle to compute the problem. I just don't see why the law of cos would go wrong here, can anybody help me out.
 
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The dog does not run in a triangle, if he did, net displacement would be zero. Try redrawing the picture. I have the exact same physics text, keep in mind northeast implies exactly 45 degrees.
 
Looks to me like the simplest way to do this problem is to reduce the first "run" to components. running 70 yards NE, Sparky runs 70(√(2)/2)= 35√(2) or about 49.5 feet east and 49.5 feet north. Now subtract the distances west and south from those to get the dogs final position. You can calculate distance and angle from his original position from that.
 
or draw your diagram to scale. For example say 1 inch = 10 m.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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