Question

http://screensnapr.com/v/zn5DFe.png
No idea.

Answer 1

use the cosine rule, u have the required lengths, the only thing missing is the angle, plug in the values into the equation and solve for \(\theta\)

Answer 2

what saso said. law 'o cosines for this one

Answer 3

a^2=b^2+c^2-2(b)(c)cosA

Answer 4

So then..

Answer 5

yes but of course you want
\[\theta\] not a length

Answer 6

9=2^2+3^2-2(2)(3)cosX

Answer 7

what you are looking for is the angle right? so you need to use
\[\theta=\cos^{-1}(\frac{b^2+c^2-a^2}{2bc})\]

Answer 8

or you could use what you wrote above to solve for
\[\cos(\theta)\] and then take the inverse cosine to get theta. amounts to the same thing

Answer 9

\(a=4\frac{1}{2} \ \ b=2 \ \ \ c=3\)

Answer 10

I got 70.5

Answer 11

One last question! Woohoo!

Answer 12

looking at the picture, u can simply tell that the angle is greater than 90... 70.5 is definitely not the right answer