System of 3 non-linear equations in 3 unknowns (how-to-solve?)

[samsonite]

This detailed screenshot http://imgbox.com/qFBxoHWd shows how I've solved, in the old past, the system by means of a MatLab function (fsolve), and I'ld like to know if a similar (short) way exists in Python.
Algorithms are listed below:

Output:
% ------ system of 3 non linear equations ----------
%
function y=f(x)
y=zeros(3,1);
rad = pi/180;
l1 = 150.210355; s1 = -38.913290; d1 = -11.185833;
l2 = 180.308440; s2 = -14.878290; d2 =  14.545555;
l3 = 223.495977; s3 =  20.492543; d3 =  -8.679444;
%---------------------------------------------------
l1 = l1 * rad; s1 = s1 * rad; t1 = tan(d1 * rad);
l2 = l2 * rad; s2 = s2 * rad; t2 = tan(d2 * rad);
l3 = l3 * rad; s3 = s3 * rad; t3 = tan(d3 * rad);
% --------------------------------------------------
y(1)= sin(x(1)+s1)-tan(x(3)+l1)*(cos(x(1)+s1)* sin(x(2))- t1* cos(x(2))); 
y(2)= sin(x(1)+s2)-tan(x(3)+l2)*(cos(x(1)+s2)* sin(x(2))- t2* cos(x(2))); 
y(3)= sin(x(1)+s3)-tan(x(3)+l3)*(cos(x(1)+s3)* sin(x(2))- t3* cos(x(2))); 
endfunction

rad = pi/180;
[x,fval,info]=fsolve(@f,[10*rad;50*rad;0*rad]) % #3 starting values in radians
la= x(1)/rad
fi= x(2)/rad
eps= x(3)/rad
% ---------------- OUTPUT ---------------------
x =

  2.6180e-001
  6.4577e-001
  5.6800e-008

fval =

  1.0311e-008
  4.9066e-008
  5.0822e-008

May I have some drifts to reach the goal? Thanks in advance

Edit. For short way I mean the use of an intrinsic function to avoid the classical procedure via an iterative method and related jacobian matrix.

[scidam]

Your code would be almost the same, if you rewrote it in Python.

from math import pi, sin, tan, cos
from scipy.optimize import fsolve

def equations(x):
    rad = pi / 180;
    l1 = 150.210355; s1 = -38.913290; d1 = -11.185833;
    l2 = 180.308440; s2 = -14.878290; d2 =  14.545555;
    l3 = 223.495977; s3 =  20.492543; d3 =  -8.679444;
    l1 = l1 * rad; s1 = s1 * rad; t1 = tan(d1 * rad);
    l2 = l2 * rad; s2 = s2 * rad; t2 = tan(d2 * rad);
    l3 = l3 * rad; s3 = s3 * rad; t3 = tan(d3 * rad);
    return (sin(x[0]+s1)-tan(x[2]+l1)*(cos(x[0]+s1)* sin(x[1])- t1* cos(x[1])), 
            sin(x[0]+s2)-tan(x[2]+l2)*(cos(x[0]+s2)* sin(x[1])- t2* cos(x[1])), 
            sin(x[0]+s3)-tan(x[2]+l3)*(cos(x[0]+s3)* sin(x[1])- t3* cos(x[1]))) 

fsolve(equations, [10 * pi / 180, 50 * pi / 180, 0])

Output:
array([2.61799394e-01, 6.45771810e-01, 1.20245548e-08])

Note: It is not pythonic way to use ; to separate statements, so, it is desirable to rewrite each assignment in a new line.

[samsonite]

Wonderful solution, scidam! Just added the output of coords in degrees. Thank you very much, and cheers

# ---------- sist_3eq.py -----------
from math import pi, sin, tan, cos
from scipy.optimize import fsolve
 
def equations(x):
    rad = pi / 180;
    l1 = 150.210355; s1 = -38.913290; d1 = -11.185833;
    l2 = 180.308440; s2 = -14.878290; d2 =  14.545555;
    l3 = 223.495977; s3 =  20.492543; d3 =  -8.679444;
    l1 = l1 * rad; s1 = s1 * rad; t1 = tan(d1 * rad);
    l2 = l2 * rad; s2 = s2 * rad; t2 = tan(d2 * rad);
    l3 = l3 * rad; s3 = s3 * rad; t3 = tan(d3 * rad);
    return (sin(x[0]+s1)-tan(x[2]+l1)*(cos(x[0]+s1)* sin(x[1])- t1* cos(x[1])), 
            sin(x[0]+s2)-tan(x[2]+l2)*(cos(x[0]+s2)* sin(x[1])- t2* cos(x[1])), 
            sin(x[0]+s3)-tan(x[2]+l3)*(cos(x[0]+s3)* sin(x[1])- t3* cos(x[1]))) 
 
print(fsolve(equations, [10 * pi / 180, 50 * pi / 180, 0]))
a= fsolve(equations, [10 * pi / 180, 50 * pi / 180, 0])
print(a[0]*180/pi,a[1]*180/pi) # coords in degrees
# ------- OUTPUT ----------
# C:\Training>python sist_3eq.py
# [2.61799394e-01 6.45771810e-01 1.20245548e-08]
# 15.00000038215297 36.99999924200017
# -------------------------