**A solution based on symmetry by Bleaug:**

**Trigonometry and calculus by Six:**

x + y + z = 360 and sinx*siny*sinz

-> f(x,y) = sin(x)*sin(y)*sin(360-x-y)

-> f(x,y) = sin(x)*sin(y)*-sin(x+y)

Now to find the critical points of f(x,y), we just need to find the partial derivative with respect to x and y and solve for 0.

f_x(x,y) =

sin(x)sin(y)(-cos(x+y)-cos(x)sin(y)sin(x+y)

solve for 0.

sin(x)sin(y)(-cos(x+y)-cos(x)sin(y)sin(x+y)=0

-> tan(x) = -tan(x+y)

Since the function is symmetric, we should get the same partial derivative for y.

-> tan(y) = -tan(y+x)

-> tan(x)=tan(y)

-> x = y or they are opposites. However, if they are opposites, the original function just becomes 0. Thus, x = y.

Now substitute in x for y in the original equation and find its critical points.

Eventually, you will get sin(3x)=0

x = 120 degrees

y = 120 degrees

z = 120 degrees

Answer (D)

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