Is that the best they can come up in 2 weeks? Complete nonsense and as others have pointed out, 100% gaslighting.
So this "triangulation" process took 15 seconds and was definitive enough to overrule the goal ump who was in perfect position?
View attachment 16873
Let A(x1, y1) and B(x2, y2) be the two observation points with bearings towards the landmark as α and β, respectively. If C(x3, y3) is the location of the landmark, the points A, B, and C form the vertices of a triangle, where:
∠CAB = θ1
and
∠CBE = θ2.
Using coordinate geometry, we can write the formula for the slope of line AC as:
tan(θ1) = (y3 - y1) / (x3 - x1) .... (1)
Similarly, for the line BC we can write:
tan(θ2) = (y3 - y2) / (x3 - x2) .... (2)
Solving equations (1) and (2) for x3 and y3, we get:
x3 = [(y1 - y2) + x2 * tan(θ2) - x1 * tan(θ1)] / [tan(θ2) - tan(θ1)]
and
y3 = [{y1 * tan(θ2) - y2 * tan(θ1)} + (x2 - x1) * tan(θ2) * tan(θ1)] / [tan(θ2) - tan(θ1)]
Wow...these ARC guys are bloody good (and fast) at trigonometry!