The triangle ABC has angles A ≈ 46.7°, B ≈ 46.7°, and C ≈ 86.6°.
To solve the triangle ABC with sides a = 5.3, b = 2.9, and c = 4.6, we can use the Law of Cosines and the Law of Sines.
First, we can use the Law of Cosines to find angle A:
![[a^2 = b^2 + c^2 - 2bc * \cos(A)]\\[\cos(A) = (b^2 + c^2 - a^2)/(2bc)]\\[\cos(A) = (2.9^2 + 4.6^2 - 5.3^2)/(2 * 2.9 * 4.6)]\\[\cos(A) \approx 0.684]\\[A \approx \cos^(-1)(0.684) \approx 46.7^\circ]]()
Next, we can find angle B using the Law of Cosines:
![[b^2 = a^2 + c^2 - 2ac * \cos(B)]\\[\cos(B) = (a^2 + c^2 - b^2)/(2ac)]\\[\cos(B) = (5.3^2 + 4.6^2 - 2.9^2)/(2 * 5.3 * 4.6)]\\[\cos(B) \approx 0.684]\\[B \approx \cos^(-1)(0.684) \approx 46.7^\circ]]()
Now, we can find angle C using the fact that the sum of the angles in a triangle is 180 degrees:
![[C = 180^\circ - A - B]\\[C \approx 86.6^\circ]]()
Therefore, the triangle ABC has angles A ≈ 46.7°, B ≈ 46.7°, and C ≈ 86.6°.