Question 515354
The dot product of those vectors is equal to (-9)(2) + (-7)(-1) = -11. This is also equal to the product of the magnitudes times the cosine of the angle in between. Hence,


*[tex \LARGE -11 = \sqrt{130}\sqrt{5}\cos{\theta}]


*[tex \LARGE \theta = \cos^{-1} \frac{-11}{\sqrt{130}\sqrt{5}} \approx 116] (degrees)