In triangle xyz, xz = yz and mz = 74°. the longest side of the triangle is: a. xz b. yz c. xy d. xm

See the attached picture to better understand the problem

we know that
if  XZ = YZ 
then
XYZ is an isosceles triangle
m ∠ Z=74°
m ∠ X= m ∠ Y
m ∠ X=(180°-74°)/2-----> m ∠ X=53°

applying the law of sines
XZ/sin 53°=XY/sin 74°
XZ=(sin 53°/sin 74°)*XY
XZ=0.83*XY
therefore
XZ < XY  or XY > XZ

the answer is
The longest side of the triangle is XY

Whose mistake? can you give more details to what the question is asking?

7.666270253x10 raised to the 18th power