This can be solved using the Law of Sines, the Law of Cosines and the general formula for the area of a triangle: Area = a * b * sin(C) / 2. This general law can be easily derived from other formulas for the area of a triangle. All three of these can be found in Wikapedia. I have a Perl script that runs an evaluator loop (in a commnd line) which is convenient for this sort of thing since it gives a compete trace. Here is the solution (you can round it yourself). Also note, that I work entirely in radians and so convert back and forth when degrees are given. I have interspersed some comments. $AC = 4.94.9 $AD = 3.83.8 $A = 80 * pi / 1801.39626340159546== Use Law of Cosides to get length of CD $CD = sqrt($AC**2 + $AD**2 - 2 * $AC * $AD * cos($A))5.65538167267988== Use Law of Sines to get unknown angles $D1 = asin($AC * sin($A) / $CD)1.02222099951733 $C1 = asin($AD * sin($A) / $CD)0.723108252476999 ($A + $C1 + $D1) * 180 / pi180== At this point the bottom left triangle ACD is completely known. $C2 = 25 * pi / 1800.436332312998582 $D2 = pi - $D12.11937165407246 $B = pi - $C2 - $D20.585888686518749 ($C2 + $D2 + $B) * 180 / pi180== Use Law of Lines to get DB and CB $DB = $CD * sin($C2) / sin($B)4.32247125729336 $BC = $CD * sin($D2) / sin($B)8.72709037836453== At this point the top right triangle BCD is completely known.== For completeness get unknowns for large triangle. $C = $C1 + $C21.15944056547558 ($A + $B + $C) * 180 / pi180 $AB = $AD + $DB8.12247125729336== Use formula for area to areas for all three triangles. $AreaACD = $AC * $AD * sin($A) / 29.16856018054366 $AreaBCD = $CD * $DB * sin($D2) / 210.4291678555696 $AreaABC = $AC * $AB * sin($A) / 219.5977280361132