Precalculus, Chapter 6, 6.1, Section 6.1, Problem 12

Using law of sines
sinA/a = sinB/b = sinC/c
A,B and C are angles and a,b and c are opposite lengths of a triangle as in the image.
According to the question we have following data.
A = 5^040' = 5.667
B = 8^015' = 8.25
b = 4.8

sinA/a = sinB/b
a = sinAxxb/sinB
a = (sin5.667)xx4.8/sin8.25 = 3.30

The addition of angles in a triangle is 180^0 .
A+B+C = 180
C = 180-5.67-8.25 = 166.08

sinA/a = sinC/c
c = asinC/sinA
c = 3.3xxsin166.08/sin5.667 = 8.04

So the answers are
Angle C = 166.08 deg
Length a = 3.3 units
Length c = 8.04 units