jeckle wrote:Seriously? No one knows how to do this? I'm kind of disappointed.

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jeckle wrote:http://tinypic.com/r/15n9g6d/6

People are more likely to help if the question is posted here, where they can see it.

For other users, the author of the image at the link wrote:. . . . .

The building lot is shaped like an isosceles triangle ABC, with AB = AC and the measure of the angle at A being 53.2 degrees. The area of the lot is one acre (which is 43,560 ft^{2}).

Find the lengths of each of the three sides, rounded to the nearest foot.

jeckle wrote:So far, I divided the triangle in half based on the two equal sides which was AB and AC. Bisection directly in the center of angle A causes 53.2 degrees to form two angles both congruent. The two congruent angles are 26.6 degrees respectfully.

The bisection also divides the isosceles to form two congruent right triangles. On the bottom of side BC or side a, it has two 90 degree angles. This allows me to find the missing side of ACB and ABC which are both congruent. The missing side is 63.4 degrees. 180-(90+26.6) = 63.4

At this point the formula to find the area of the triangle is A = 1/2 * a * b * SinC

I'm getting confused. Is "A" here an angle, a vertex, or a side? (I think it's

supposed to be a side, but you've got "a", "b", and "c" for that...?)

Here's another approach:

Split the triangle like you said, and reorient (do a new drawing) where BC is the base (that is, where it's at the bottom of the triangle, rather than being to the right). Label the height (from A to the middle of BC) as "h" for the height, and half the base as "x" (so we've got a handy name for it).

The area of the whole triangle is (1/2)(base)(height) = (1/2)(2x)(h) = xh = 43,560.

Using just the half-triangle, we get x = (AB)sin(A) and h = (AB)cos(A).

Multiply these last two expressions for x and h, set equal to the previous expression for xh, and solve for the value of AB. And so forth.