What is the area of △FGH to the nearest tenth of a square meter? The image is of a triangle GHF with base GH length 2m, FG is 2.5 m and angle FGH is 121 degrees.

First, we are going to use the law of cosines to find the length of the line segment FH:
[tex]FH= \sqrt{2.5^{2}+2^{2}-(2)(2.5)Cos(121)} [/tex]
[tex]FH= \sqrt{2.5^{2}+2^{2}-5Cos(121)} [/tex]
[tex]FH=3.2[/tex]

Next, we are going to use the semi-perimeter formula: [tex]s= \frac{GH+FG+FH}{2} [/tex]
[tex]s= \frac{2+2.5+3.2}{2} [/tex]
[tex]s= \frac{7.7}{2} [/tex]
[tex]s=3.9[/tex]

Now that we have the semi-perimeter of our triangle, we can find its area using Heron's formula:
[tex]A= \sqrt{s(s-GH)(s-FG)(s-FH)} [/tex]
[tex]A= \sqrt{3.9(3.9-2)(3.9-2.5)(3.9-3.2)} [/tex]
[tex]A= \sqrt{3.9(1.9)(1.4)(0.7)} [/tex]
[tex]A=2.7m^{2}[/tex]

We can conclude that the area of the triangle GHF is 2.7 [tex]m^{2}[/tex].