- We know that the triangles circumference is at 4.8 meters,
- one angle is at 60°
- and that this angle's opposite side is 1.7m in length.
How would I go about solving for this triangles area?
I can't believe we chose English to be the lingua franca.
- ted sLv 71 month ago
what do YOU mean by " triangles circumference " ???
- az_lenderLv 71 month ago
I guess you're saying the triangle's PERIMETER is 4.8 meters. (?)
I'll try the law of cosines:
(1.7)^2 = x^2 + (4.8 - 1.7 - x)^2 - 2x(3.1 - x)*cos(60) =>
2.89 = x^2 + 9.61 - 6.2x + x^2 - 3.1x + x^2 =>
0 = 3x^2 - 9.3x + 6.72 =>
the other two sides are 1.953 m and 1.147 m.
Now you can do the area as
(1/2)(1.953 m)(1.147 m)*sin(60) = 0.970 m^2.