In ΔPQR, q = 420 inches, p = 400 inches and ∠P=64°. Find all possible values of ∠Q, 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 the side of length c, the following equation holds true:

c² = a² + b² - 2ab cos(γ)

In this case, we have a triangle PQR with sides p = 400 inches, q = 420 inches, and an angle P = 64° opposite the side of length p. We want to find the angle Q, which is opposite the side of length q.

We can rearrange the Law of Cosines to solve for cos(γ):

cos(γ) = (a² + b² - c²) / (2ab)

Substituting the given values into this equation gives us:

cos(Q) = (p² + r² - q²) / (2pr)

We don't know the length of side r, but we can express it in terms of p and Q using the Law of Cosines:

r² = p² + q² - 2pq cos(P)

Substituting this into our equation for cos(Q) gives us:

cos(Q) = (p² + (p² + q² - 2pq cos(P)) - q²) / (2p * sqrt(p² + q² - 2pq cos(P)))

Solving this equation for Q will give us the possible values of ∠Q.

Please note that this equation may have two solutions, since cos(γ) = cos(360° - γ) for any angle γ. This means that if Q is a solution, then 360° - Q is also a solution.

Finally, round your answers to the nearest degree.