Does anybody know it ??

Work Shown:

The given values are

Through the Pythagorean trig identity, we can find

I'm skipping these steps, but let me know if you need me to go over them.

Since u and v are in quadrant 3, this means both sine and cosine are negative here

Using those four items, we can then say the following:

cos(v-u) = cos(v)cos(u) + sin(v)sin(u)

cos(v-u) = (-8/17)*(-5/13) + (-15/17)*(-12/13)

cos(v-u) = 40/221 + 180/221

cos(v-u) = (40 + 180)/221

cos(v-u) = 220/221

This is reduced as much as possible because the GCF of n and n+1 is always 1.