Trigonometry Problem - Solution

A new solution for previous trigonometry problem.

In previous post, I have provided a solution for the problem by triangle area formula, however it seen complicated.

Recently, I have thought up a new solution for it.

Consider the triangle in Cartesian coordinate system, using linear equation for developing the equations for lines AB, AD, CE and BC.

First, assume length of AC is 1

line AB:  y = x tan (80)

line AD:  y = x tan (60)

line CE:  y =  x tan (110) - tan (110)

line BC:  y = x tan (100) - tan (100)

Point E is intersection of lines AB and CE;

x tan (80) = x tan (110) - tan (110)

x tan (80) - x tan (110) = - tan (110)

x tan (110) - x tan (80) = tan (110)

x [tan (110) - tan (80)] = tan (110)

          x = tan (110) / [tan (110) - tan (80)]

          x = 0.32635

using line AB equation;

          y = 0.32635 tan (80)

          y = 1.8508

Point E is (0.32635, 1.85083)

With same method; find for coordinate of D

Point D is (0.766044,1.32683)

Then, find slope of DE by two points;

The slope DE = -1.191753

          tan (d) = -1.191753

                  d = -50 degree or 50 degree

Angle of line EC to horizontal is 70 degree

Then angle e is 70 - 50

thus, answer e = 20 degree