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