To find the value of angle ∠I in ΔHIJ, we can use the Law of Cosines, which states that for a triangle with sides a, b, and c, and angle A opposite side a, we have:
c^2 = a^2 + b^2 - 2ab cos(A)
In this case, we know the lengths of sides HI and HJ, and the measure of angle H. We want to find the measure of angle I, so we will use the Law of Cosines to solve for side HJ:
HJ^2 = HI^2 + HJ^2 - 2(HI)(HJ)cos(∠I)
Simplifying this equation, we get:
2(HI)(HJ)cos(∠I) = HI^2 + HJ^2 - HJ^2
2(HI)(HJ)cos(∠I) = HI^2
cos(∠I) = HI^2 / 2(HI)(HJ)
cos(∠I) = HI / 2(HJ)
cos(∠I) = 460 / (2 * 550)
cos(∠I) = 0.41818181818181815
Now, we can use the inverse cosine function (cos^-1) to find the possible values of ∠I:
∠I = cos^-1(0.41818181818181815)
∠I ≈ 65.6°
Note that the Law of Cosines can give two possible values for an angle in a triangle, but in this case, we only get one valid value since the given angle H is obtuse (greater than 90°) and the other possible angle value obtained by the Law of Cosines would be greater than 180°, which is not possible. Therefore, the only possible value of ∠I in this triangle is approximately 65.6° to the nearest degree.