In ΔEFG, g = 530 cm, f = 460 cm and ∠F=139°. Find all possible values of ∠G, to the nearest degree.

To solve this problem, we will use the Law of Cosines. The Law of Cosines states that for any triangle with sides of lengths a, b, and c, and an angle γ opposite side c, the following equation holds true:

c² = a² + b² - 2ab cos(γ)

In this case, we have a triangle ΔEFG with sides g = 530 cm, f = 460 cm, and ∠F = 139°. We want to find ∠G, so we'll rearrange the Law of Cosines to solve for cos(γ):

cos(γ) = (a² + b² - c²) / (2ab)

Substituting the given values:

cos(G) = (f² + g² - e²) / (2fg)

We don't know the length of side e, but we can express it in terms of f, g, and ∠F using the Law of Cosines:

e² = f² + g² - 2fg cos(F)

Substituting the given values:

e² = 460² + 530² - 2460530 cos(139°)

Calculate e² and take the square root to find e.

Then substitute f, g, and e into the rearranged Law of Cosines to find cos(G). Finally, use the arccos function to find ∠G:

G = arccos[(f² + g² - e²) / (2fg)]

Remember that the arccos function will give you an angle in radians, so you'll need to convert to degrees by multiplying by 180/π. Round your answer to the nearest degree.