Map question involving vectors (find the angle)

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Homework Statement


Instructions for finding a buried treasure include the following: Go 66.0 paces at 256deg, turn to 140deg and walk 125 paces, then travel 100 paces at 169deg. The angles are measured counterclockwise from an axis pointing to the east, the +x direction. Determine the resultant displacement from the starting point. Enter the distance (without units) and the angle relative to the positive x-axis.

Homework Equations

The Attempt at a Solution


I already figured out the displacement which is 213 paces but i thought the angle could be found using the two components (x and y), 35.4 and -210. Please help because I'm wrong!
 
on Phys.org
so if you do ##\tan\theta=\frac{-35.4}{210}## the angle you get isn't the correct answer? I suspect that the answer is given in positive number , which other positive angle has the same tangent as that negative angle?
 
Delta² said:
so if you do ##\tan\theta=\frac{-35.4}{210}## the angle you get isn't the correct answer? I suspect that the answer is given in positive number , which other positive angle has the same tangent as that negative angle?
I tried that as well but it’s apparently still wrong
 
Delta² said:
I didn't check your answer for the displacement, is 213 paces correct?
Yes it’s 213 paces
 
Delta² said:
What's the answer key for the angle?
It unfortunately doesn’t say, its an online where it tells me whether I’m right or wrong
 
Delta² said:
So neither -9.568 degrees or 350.432 degrees is the correct answer?
Could you explain to me why it would be 350.432 degrees?
 
Delta² said:
Trigonometry formulas say that the tangent of angle ##-\theta## is equal to the tangent of angle ##2\pi-\theta## or ##360-\theta## in degrees.
Well i just tried 3.50 x 10^2 degrees (since i can only carry 3 significant digits) and unfortunately still a no :(,
 
Delta² said:
Well I don't know what else, maybe you calculated the x and y as y,x, try ##\tan\theta=-\frac{210}{35.4}## which leads to ##\theta=-80.4## or ##\theta=279.6##
Thank you so much for your help I really appreciate it!
 
For the counterclockwise angle from the positive-x-axis, [itex]\theta=\tan^{-1} \left(\frac{R_y}{R_x}\right)[/itex],
with the rule of thumb to add [itex]180^\circ[/itex] if [itex]R_x<0[/itex] (since [itex]\tan^{-1}[/itex] returns a value between [itex]-90^\circ[/itex] and [itex]+90^\circ[/itex]).