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 ).
Group
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
You can also select a web site from the following list
How to Get Best Site Performance
Select the China site (in Chinese or English) for best site performance. Other MathWorks country sites are not optimized for visits from your location.
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 ).