[smileshaikh]

All right, so basically, I've tried to work this problem out a billion times. I keep getting an answer of 16352.98348but the answer on the homework question that the teacher gave us is 16,060 m^2... what's the deal?

Question:

"Find the area of a triangular-shaped field with sides of 199.8m and 181.4m, and the included angle between them measuring 62.40 degrees."

[ebaines]

If we take the long side as the base of the triangle (base = 199.8m), then the height of the trangle as measured from the base is 181.4m times the sine of 62.4 degrees. The area of the triangle is one half base length times height, so you get:

A = 199.8m x 181.4m x sin(62.4 degrees)


I get 16059.66 m^2, which rounds to 16060 m^2

[smileshaikh]

Thank you very much for your help. I'm curious though, just from how the question is stated, how did you know to use the longer side as the base? Were we to just assume this was a right triangle? Somehow I thought I was working with two right triangles... combined.

[ebaines]

It's not a right triangle. The area of any triangle is 1/2 base times height - doesn't matter whether it's a right triangle (one angle = 90 degrees), acute (all three angles less than 90 degrees), or obtuse (one angle > 90 degrees).

We could have chosen the other side as the base, and the height as measured from that side would be 199.8m times sin(62.4 degrees). So the area would be

A = 1/2 base times height = 1/2 x 181.4m x (199.8m x sin(62.4))

which is the same answer. See the attached drawing which shows the two cases.