Problem 49933. Splitting Triangle - Problem the second

I actually solved this by solving for the area of a triangle given a side and the adjacent angles, then set one of them to 60 degrees and simplified.
Given : side c and angles α and β.
We know area = c b sinα / 2.
We can calculate b through the law of sines:
b = c sinβ / sinγ
We also know α + β + γ = 180, so γ = 180 - (α + β).
But sin is even symmetric about 90, so sinγ = sin(α + β).
We have: area = ½c² sinα sinβ / sin(α + β).
By the angle sum formula for sine, area = ½c² sinα sinβ / (cosα sinβ + sinα cosβ).
Dividing numerator and denominator by sinα sinβ yields: area = ½c² / (cosα /sinα + cosβ / sinβ ).
But cos x / sin x = cot x, so: area = ½c² / (cotα + cotβ ).