Displacement of a vector (Really simple)

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Homework Help Overview

The problem involves a ship's displacement after being blown off course by a storm, requiring the determination of both the distance and direction to return to its original destination. The subject area includes vector analysis and trigonometry.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of angles related to the displacement vector, questioning whether the angle found is the correct one to use for the problem's requirements. There is mention of complementary and supplementary angles in relation to the triangle formed by the ship's path.

Discussion Status

The discussion is ongoing, with participants providing insights into the geometric interpretation of the problem. There is an acknowledgment of the need to clarify the angle's definition in the context of the problem, but no consensus has been reached on the correct angle to use.

Contextual Notes

Participants are navigating the definitions of angles in relation to the axes provided in the problem, indicating potential confusion regarding the terms "complementary" and "supplementary" in this context.

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[SOLVED] Displacement of a vector (Really simple)

Homework Statement


A ship sets out to sail to a point 106 km due north. An unexpected storm blows the ship to a point 128 km due east of its starting point. (a) How far (in km) and (b) in what direction (as an angle from due east, where north of east is a positive angle) must it now sail to reach its original destination?


Homework Equations





The Attempt at a Solution


I have (a) = 166.1926593 which is correct and to find (b) I'm just using tan^(-1)(106/128) = 39.6290053 but this isn't correct? Am I supposed to be taking the complement of this angle?
 
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What you've solved for is the angle inside the right angled triangle, from what I gather. If you take for example, the up/down and left/right axes to be the directions and draw them in on the triangle with the new starting point (128 east of the original starting point) as the origin, that should help. The problem tells you to take the angle from the east, with north of east as positive, so find the angle from the east axis to the hyp. of the triangle. That is the supplementary angle to the one you have found and it should be the answer you are looking for. :)
 
Well the opposing angle would be 50.3709947 which is also incorrect. I would have thought the angle would have just been 90 seeing as its only involving direct N and direct E.
 
You should be looking for the supplementary angle, which is 180 degrees-theta. What you've found is the complementary angle.
 
Ah I gotcha thanks
 

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