In triangle xyz, side xy measures 6 inches, side yz measures 8 inches, and side xz measures 10 inches. list the angles of the triangle in order from largest angle to smallest angle

we have that

In triangle XYZ,
side XY measures 6 inches,
side YZ measures 8 inches,
side XZ measures 10 inches

let
a> side XY> 6 in
b> side YZ> 8 in
c> side XZ> 10 in
A> angle YZX
B> angle YXZ
C> angle XYZ
we know that
we can apply the Cosine Law to find the angles of the triangle.

find angle C
c²=a²+b²-2ab*cosC
cos C=(a²+b²-c²)/(2ab)>  cos C=(6²+8²-10²)/(2*6*8)
cos C=0
C=arc cos (0)> C=90°

find angle B
b²=a²+c²-2ac*cosB
cos B=(a²+c²-b²)/(2ac)>  cos B=(6²+10²-8²)/(2*6*10)
cos B=0.60
B=arc cos (0.60)> B=53.12°

A+B+C=180> A=180-(B+C)> A=180-(90+53.12)
A=36.88°

the answer is
the angles of the triangle in order from largest angle to smallest angle are
C=90°> angle XYZ
B=53.12°> angle YXZ
A=36.88°> angle YZX