Can someone me with this math question ? ?

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.

a

b

d

step-by-step explanation:

b

step-by-step explanation:

u take $75 minus $60.the anwer/original($75)multiply by 100