Hi Ben,
I reproduced the diagram you sent
I divided the lot into three triangles and labeled the vertices.
I am going to find the lengths of the two lines I added using the law of cosines, so I need to express the angle measurements using decimal fractions of degrees so that I can use my calculator to calculate the cosines.
25' is ^{25}/_{60} = 0.417^{o} and 48' is ^{48}/_{60} = 0.8^{o}
Thus the measure of angle A is 76.417^{o} and the measure of angle D is 44.8^{o} .
The law of cosines, applied to triangle ABE gives
BE^{2} = AB^{2} + EA^{2}  2ABEA cos(A)
= 322^{2} + 113^{2}  2 322 113 cos( 76.417^{o} )
= 103684 + 12769  72772 0.23485
= 99362.225
Ant thus BE = √ 99362.225 = 315.21 feet.
In a similar fashion I found EC = 318.61 feet
Now I can use Heron's formula to calculate the areas of the three triangles.
For example, for triangle ABE let a = 113, b = 322 abd c = 315.21, then
s = (a + b + c)/2 = 375.105 and area = Sqrt[s (s  a)(s  b)(s  c)] = 17683.9 square feet.
Similarly the area of triangle BCE is 2314.31 square feet and the area of triangle CDE is 9854.33 square feet. Thus the area of your lot is
17683.9 + 2314.31 + 9854.33 = 29852.5 square feet.
There are 43560 square feet in an acre so your lot has an area of
^{29852.5}/_{43560} = 0.69 acres
Harley
