Three charges on corner of triangle

Suppose three charges are at the corner of a triangle.
   A
  /  \
 /    \
B -- C


A = 4 C
B = 2 C
C = 2 C

distance from A to B = 3m
distance from A to B = 3m
distance from B to C = 1.2m
Note: A is located equidistant from B and C

Calculate the electric field at point A. Include direction and magnitude of the electric field.

Fix the coordinate axes with origin at A.

Angle BAC = 2 θ
From tan θ = 0.6/3 we get θ = 11.3°

Force due to B at A is at angle 11.3°
Force due to C at A is at angle 168.7°

The horizontal components cancel each other and the
Vertical components add.

Force at A due to B is kq1q1/r² = 9e9* 2*4/3² = 8.e+9 N
Its component in the vertical direction is 8.e+9* sin11.3°
= 1.57e+9 N

Adding the force due to the other charge in this direction we get
net force = 2*1.57e+9 N = 3.14e+9 N
The direction is vertically up .
============================