How Does Walking West Affect Angles in Triangle SCT?

  • Thread starter Thread starter ane
  • Start date Start date
AI Thread Summary
Mike walks 100 meters west from a position directly north of a tree, forming an angle of 82 degrees between his starting point, current position, and the tree. His starting point is designated as vertex S, his current position as C, and the tree as T, with angle SCT confirmed as 82 degrees. Participants in the discussion agree with Mike's interpretation of the scenario. There is no indication of confusion regarding the angle measurement. The consensus is that Mike's understanding of the triangle's angles is correct.
ane
Mike is standing due north of a tree. He walks due west 100m and realizes that the angle between his start position, this current position and the tree is 82 deg. Assuming his start position is the vertex S , his current position is C and the tree is T. The angle SCT is 82 deg. Am I right?
 
Physics news on Phys.org
ane said:
Mike is standing due north of a tree. He walks due west 100m and realizes that the angle between his start position, this current position and the tree is 82 deg. Assuming his start position is the vertex S , his current position is C and the tree is T. The angle SCT is 82 deg. Am I right?
That's how I read it.
 
me too. seems clear enough
 
ane said:
Mike is standing due north of a tree. He walks due west 100m and realizes that the angle between his start position, this current position and the tree is 82 deg. Assuming his start position is the vertex S , his current position is C and the tree is T. The angle SCT is 82 deg. Am I right?

Yes, you are right. Why are you in doubt?
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top