Why is this certain angle 20 degrees?

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The discussion centers on understanding why the angle VB/A is 20 degrees in the context of a problem involving two ships. Participants express confusion about how the angle from ship A to ship B, which is given as 20 degrees, translates to the angle VB/A in the solution diagram. It is clarified that the angle remains 20 degrees due to the properties of parallel lines cut by a transversal, leading to equal alternating interior angles. The importance of recognizing the geometric relationships in the diagram is emphasized, as assuming otherwise could hinder problem-solving. Ultimately, the angle is crucial for solving the problem correctly.
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Homework Statement


Problem #199
http://img88.imageshack.us/img88/8008/scan0001vd.jpg
Solution
http://img21.imageshack.us/img21/2131/199hg.jpg

Homework Equations


Why is the angle from VB/A 20 degrees from the solution diagram? It would seem that I had to know that direction of VB/A had the same angle as VA in terms of the geometry (if two parallel lines are cut by a transversal, its alternating interior angles are equal). I just don't see how you can assume that.


The Attempt at a Solution


Since drawing a triangle is the first part, I don't have any "attempt" at it yet.
 
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I'm not sure what you're asking.

You are given that the bearing is 20 degrees. i.e. A takes a bearing of B and sees it is 20 degrees East of North, thus theta is 20 degrees.
 
I saw that 20 degrees was given, I just don't see how the angle is 20 degrees on the solution diagram. Ship A observes ship B at 20 degrees, but how is VB/A also 20 degrees down from horizontal? Sorry for the confusion!
 
elementG said:
I saw that 20 degrees was given, I just don't see how the angle is 20 degrees on the solution diagram. Ship A observes ship B at 20 degrees, but how is VB/A also 20 degrees down from horizontal? Sorry for the confusion!

It's been while, sorry, VB/A represents what part of the diagram?

I'd assumed we only care about angle theta, which is 20.
 
VB/A comes off the head of VA. I just don't see how its 20 degrees when VB/A and VA are connected as seen on the solutions diagram.
 
elementG said:
VB/A comes off the head of VA.
Sorry, I hadn't looked at the second diagram.
elementG said:
I just don't see how its 20 degrees when VB/A and VA are connected as seen on the solutions diagram.

Well, the solution triangle is just a rearrangement of the starting configuration. The two vectors start off at 20 degrees, why would that change?
 
Oh, I guess I made the wrong assumption. I was assuming the angle that VB/A made was not necessarily 20 degrees. I guess I'm confused (a little bit) still is because I can't see it geometrically. Like say for instance, I'm still on the assumption that the angle is not 20 degrees for VB/A and I label as an unknown, how would I geometrically prove that the angle is 20 degrees?
 
elementG said:
Oh, I guess I made the wrong assumption. I was assuming the angle that VB/A made was not necessarily 20 degrees. I guess I'm confused (a little bit) still is because I can't see it geometrically. Like say for instance, I'm still on the assumption that the angle is not 20 degrees for VB/A and I label as an unknown, how would I geometrically prove that the angle is 20 degrees?

You would not be able to solve the problem. You're given the angle because you need it.
 

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