Let the length of each side opposite to vertex A, B and C be a, b and c respectively, and the diameter of circumscribed circle of ΔABC be D.
The law of sines will be written as follows: a/sinA = b/sinB = c/sinC = D ··· Eq.1
So we have: c = D·sinC We also have the altitude h, drawn from vertex A to the opposite side BC: h = c·sinB = (D·sinC)·sinB = D·sinB·sinC
Therefore,we have the area of ΔABC S: S = (a·h)/2 = (a·D·sinB·sinC)/2 ··· Eq.2
From Eq.1 we also have: D = a/sinA ··· Eq.3
Plug Eq.3 into Eq.2. S = {a·(a/sinA)·sinB·sinC}/2 = (a²·sinB·sinC)/(2sinA)**
Answer: Area of ΔABC = (a²·sinB·sinC)/(2sinA)
** If b,c are also given or evaluated, the expression above would be written as follows: Area of ΔABC = (a²·sinB·sinC)/(2sinA) = (b²·sinC·sinA)/(2sinB) = (c²·sinA·sinB)/(2sinC)