Étant donné deux points Aet B, trouvez l'angle d'une ligne AOà l'autre BOautour du point Ooù se Otrouve l'origine ( (0,0)). De plus, l'angle peut être positif ou négatif selon la position des points (voir exemples). L'entrée sera des points Aet B, et peut être donnée sous n'importe quelle forme pratique. La sortie sera l'angle en degrés (mais elle est positive si elle AOest tournée dans le sens antihoraire autour de l'origine pour obtenir BOet négative si elle est tournée dans le sens horaire). Si l'angle est de 180 degrés, vous pouvez renvoyer une sortie négative ou positive. De même, l'angle peut être la version positive ou négative du même angle ( 90 degest égal à -270 deg). Exemples:

• Entrée: A(5,5) B(5,-5)Sortie: -90( AOest tourné -90degrés pour obtenir BO).

• Entrée: A(5,-5) B(5,5)Sortie: 90( AOest tourné 90degrés pour obtenir BO).

C'est le code-golf , donc le code le plus court en octets gagne!

Pyth, 11 octets

.t-FPM.jMQ6

Manifestation

L'entrée est donnée au format:

[[Bx, By], [Ax, Ay]]

Si l'on souhaite que A vienne en premier, cela peut être modifié pour 1 octet.

Explication:

.t-FPM.jMQ6
               Implicit: Q = eval(input())
      .jMQ     Convert input pairs to complex numbers.
    PM         Take their phases (angles in the complex plane).
  -F           Take the difference.
.t        6    Convert to degrees

TI-BASIC, 13 octets

Pour les calculatrices de la série TI-83 + / 84 +.

Degree
Input Y
min(ΔList(R►Pθ(Ans,∟Y

Pour utiliser ce programme, entrez la liste {x1,x2}via la variable Ans et {y1,y2}à l'invite.

CJam, 14 octets

q~::ma:-P/180*

Il s'agit d'un programme complet qui lit l'entrée à [[Ax Ay] [Bx By]]partir de STDIN.

Essayez-le en ligne dans l' interpréteur CJam .

Comment ça fonctionne

q~             e# Read and evaluate all input.
  ::ma         e# Replace each point (x, y) with atan2(x, y).
               e# This returns its angle with the positive y axis, measured clockwise.
      :-       e# Compute the difference of the two resulting angles.
               e# This returns the angle between the points, measured counter-clockwise.
        P/180* e# Divide by Pi and multiply by 180 to convert to degrees.

Minkolang 0.9 , 112 octets

I really want to implement trig functions as built-ins now...but this was fun! (Caveat: this outputs the positive angle difference, not the signed angle difference. Given my limitations, I think that's justified.)

4[n]0c2c*1c3c*+r4[2;1R]r+1R+0g*12$:;$:8[0ci2*3+d1R;0g$:1i1+[i2*1+d1+$:*]*]$+'3.141592654'25*9;$:$:12$:r-66*5**N.

Try it here.

Explanation

Je posterai une explication plus complète si quelqu'un le veut, mais l'essentiel est:

4[n]                                    Take in 4 integers from input
0c2c*1c3c*+                             dot product
r4[2;1R]r+1R+0g*12$:;                   magnitudes of vectors
$:                                      dot product divided by magnitudes (z)
8[0ci2*3+d1R;0g$:1i1+             *]    Taylor series for arccos
                     [i2*1+d1+$:*]      In particular, the coefficient (1/2 * 3/4 * ...)
$+                                      Add them all up!
'3.141592654'25*9;$:$:                  Divide by pi for converting to degrees
12$:r-                                  Subtract from 1/2 - I now have arccos(z)
66*5**                                  Convert to degrees
N.                                      Output as number and stop.

Mathematica, 22 bytes

{-1,1.}.ArcTan@@@#/°&

Example:

In[1]:= {-1,1.}.ArcTan@@@#/°&[{{5,5},{5,-5}}]

Out[1]= -90.

In[2]:= {-1,1.}.ArcTan@@@#/°&[{{5,-5},{5,5}}]

Out[2]= 90.

Javascript, 66 bytes

let f=(a,b)=>(Math.atan2(b.y,b.x)-Math.atan2(a.y,a.x))*180/Math.PI;

demo

Julia, 18 25 bytes

f(A,B)=angle(B/A)/pi*180

This assumes that "any convenient form" already allows for A and B to be given as complex numbers. Then, the complex number arithmetic does all the heavy lifting.

Edit: converted snippet to function. 18 byte version only works in the Julia REPL.

Python 2.7, 73 Bytes

from math import*
f=lambda A,B:degrees(atan2(B[1],B[0])-atan2(A[1],A[0]))

Test:

f((5,5),(5,-5)) #-90.0
f((5,-5),(5,5)) #90.0

Octave, 43 bytes

f=@(a,b)(cart2pol(b)-cart2pol(a))(1)*180/pi

Input/Output:

octave:40> f([5,5],[5,-5])
ans = -90

octave:41> f([1,0],[0,1])
ans = 90

CJam, 15 bytes

l~ma@@ma-P/180*

Thought I'll get in the CJam game as well. Try it online. Input is in form of bx by ax ay. Unfortunately, this is the shortest method of doing this challenge without copying Dennis' answer.

TeaScript, 28 bytes

I really should of implemented trig functions...

$.atan2(_[3]-y,z-x)*180/$.PI

Try it online input is a.x a.y b.x b.y

Explanation

$.atan2(       // Arc Tangent of...
    _[3] - y,  // 4th input - 2nd input
       z - x,  // 3rd input - 1st input
) * 180 / $.PI // Converts rad -> deg

Ruby, 64, 58 bytes

a=->(b){b.map{|c|Math.atan2(*c)}.reduce(:-)*180/Math::PI}

Usage

a.call [[5, 5], [5, -5]] # => -90.0
a.call [[5, -5], [5, 5]] # => 90.0

JavaScript, 49 bytes

(a,b)=>((c=Math.atan2)(...b)-c(...a))/Math.PI*180

Input is taken in form: [aY, aX], [bY, bX] (notice the reversed x/y)

Simplex v.0.7, 13 bytes

I'm glad I added mathrelations :D Unfortunately, I cannot take pointwise input. So, I input each point as a separate number (Ax, Ay, Bx, By). (I used this as a resource.)

(iRi~^fR)2LSo
(       )2    ~~ repeat inner twice
 iRi          ~~ take two chars of input (x,y)
    ~         ~~ switch top 2 on stack
     ^f       ~~ apply atan2 on (y,x)
       R      ~~ go right
          L   ~~ go left
           S  ~~ subtract result
            o ~~ output as number

I can save a char if I can take input as (Ay, Ax, By, Bx):

(iRi^fR)2LSo

C, 88 bytes

#include<math.h>
typedef double d;d g(d x,d y,d a,d b){return atan2(b-y,a-x)*180/M_PI;}

Requires compiling with GCC to take advantage of M_PI being defined in math.h as a part of GCC's built-in math constants. Try it online - since ideone doesn't use GCC (apparently), an additional few bytes are needed for enough digits of π to be accurate.