Question

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Answer 1

Your questions are always incomplete.

Answer 2

law of cosines for this one
which angle do you want to find first?

Answer 3

ok good idea since it is the biggest one
the law of cosines tells you
\[a^2=b^2+c^2-2b\cos(A)\] solving for \(\cos(A)\) gives you
\[\cos(A)=\frac{b^2+c^2-a^2}{2bc}\]

Answer 4

plug in the numbers and get
\[\cos(A)=\frac{3^2+1.2^2-4^2}{2\times 3\times 1.2}\]

Answer 5

i get \(-.772\) approximately, making \[A=\cos^{-1}(-.772)\]

Answer 6

easiest to keep the numbers in the calculator and do it all at once

Answer 7

http://www.wolframalpha.com/input/?i=arccos%28%283^2%2B1.2^2-4^2%29%2F%282*+3*1.2%29%29
i get about \(140.6\) degrees

Answer 8

then ditch the law of cosines for the other one and revert to the law of sines
it is easier
third is easiest, make sure they add up to 180

Answer 9

sure

Answer 10

by the law of sines
\[\frac{\sin(B)}{3}=\frac{\sin(140.6)}{4}\] so
\[\sin(B)=\frac{3\sin(140.6)}{4}\] take the arcsine of the result to get your answer

Answer 11

looks like about \(31.53\)
http://www.wolframalpha.com/input/?i=arcsin%28 \frac{3\sin%28140.6%29}{4}%29

Answer 12

i am going to stick with my answer because if the other one is 61 then the total is already larger than 180 before you even find the third angle!

Answer 13

yw

Answer 14

umm