Question

In ∆ABC, AB = 4.12, BC = 4, AC = 5. What is m C?

Answer 1

you got a picture? looks like this should be a law of cosines problem
does that seem right?

Answer 2

Law of cosines is correct

Answer 3

no theres no pictures and yes thats what it
is

Answer 4

lets draw one and label it

Answer 5

ok idk how to tho

Answer 6

ok now what

Answer 7

my picture stinks, because 4 is larger than 5, but mo matter at least we have something to refer to

Answer 8

\[c^2=a^2+b^2-2ab\cos(C)\] we are looking for \(C\)

Answer 9

no its fine

Answer 10

solving for \(\cos(C)\) you get
\[\cos(C)=\frac{a^2+b^2-c^2}{2ab}\]

Answer 11

now we put in the numbers you are given

Answer 12

\[\cos(C)=\frac{4^2+5^2-(4.12)^2}{2\times 4\times 5}\]

Answer 13

ok

Answer 14

and so \(C\) is the arccosine of that number
we need a calculator to find it

Answer 15

i just plugged it in directly to wolfram and it gave me this
http://www.wolframalpha.com/input/?i=arccos%28%285^2%2B4^2-4.12^2%29%2F40%29

Answer 16

\[\frac{ 16+25-(4.12)^2 }{40 ? }\]

Answer 17

yes that is the cosine of the angle
you want the angle itself so you want
\[\cos^{-1}\left(\frac{ 16+25-(4.12)^2 }{40 }\right)\]

Answer 18

typed it in to wolfram and got about \(53.08\)