## Lukecrayonz 4 years ago http://screensnapr.com/v/zn5DFe.png No idea.

1. sasogeek

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 $$\theta$$

2. anonymous

what saso said. law 'o cosines for this one

3. Lukecrayonz

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

4. Lukecrayonz

So then..

5. anonymous

yes but of course you want $\theta$ not a length

6. Lukecrayonz

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

7. anonymous

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

8. anonymous

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

9. sasogeek

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

10. Lukecrayonz

I got 70.5

11. Lukecrayonz

One last question! Woohoo!

12. sasogeek

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

13. sasogeek