Can someone help me with this math question please??

Answer:
3*2 =6.2*2+4*2-2*6.2*4 cos(C)
just simplify
Step-by-step explanation:
The triangle ABC has side lengths a, b and c. Draw AP perpendicular to BC, let h be the length of AP and x be the length of BP, then the length of PC is a - x.
Write Pythagoras' theorem for the two right triangles
x2 + h2 = c2
h2 + (a - x)2 = b2
I need one further equation
cos(C) = (a - x)/b or
a - x = b cos(C)
and thus
x = a - b cos(C)
Expand the second equation to get
h2 + a2 - 2ax + x2 = b2
Using the first equation replace x2 + h2 by c2 then using the third equation replace x by a - b cos(C). Simplify to get
c2 = a2 + b2 - 2ab cos(C)
c=3
b=4
a=6.2
there for: 3*2 =6.2*2+4*2-2*6.2*4 cos(C)
This is the Law of Cosines. If you know the side lengths a, b and c you can find cos(C) and hence the measure of the angle C.