Solve the triangle with the given parts.
\[
a=21.5 m, b=15.3 m, c=6.3 m
\]
What is the degree measure of angle A?
(Round to the nearest tenth as needed.)

Step 1 ：We are given a triangle with sides of lengths \(a = 21.5 m\), \(b = 15.3 m\), and \(c = 6.3 m\). We are asked to find the degree measure of angle A.

Step 2 ：We can use the Law of Cosines to solve for angle A. The Law of Cosines states that for any triangle with sides of lengths a, b, and c, and an angle A opposite side a, the following equation holds: \(a^2 = b^2 + c^2 - 2bc \cos(A)\)

Step 3 ：We can rearrange this equation to solve for \(\cos(A)\), and then use the inverse cosine function to find the measure of angle A: \(\cos(A) = \frac{b^2 + c^2 - a^2}{2bc}\) and \(A = \cos^{-1}\left(\frac{b^2 + c^2 - a^2}{2bc}\right)\)

Step 4 ：Substituting the given values into this equation, we find that \(\cos(A) = -0.9776429090154579\)

Step 5 ：Taking the inverse cosine of this value, we find that \(A = 2.9297394127210805\) radians

Step 6 ：Converting this to degrees, we find that \(A = 167.9\) degrees

Step 7 ：Final Answer: The degree measure of angle A is \(\boxed{167.9}\) degrees