The linear algebra tells us that the angle between two vectors is given in proportion to the point product between the two vectors and inversely proportional to the product between the magnitudes of both vectors, that is:
[tex]cos\theta = \frac{u\cdot v}{|u||v|}[/tex]
Consider our vecores as
[tex]u = 20\hat{i} + 40\hat{j}[/tex]
[tex]v = -2\hat{j} + 9\hat{k}[/tex]
The dot product of these two vectors would be
[tex]u\cdot v = (20)(0)+(40)(-2)+(9)(0) = -80[/tex]
The magnitude of the two vectors would be
[tex]|u| = \sqrt{20^2+40^2} = 20\sqrt{5}[/tex]
[tex]|v| = \sqrt{(-2)^2+9^2} = \sqrt{85}[/tex]
Now the angle will be
[tex]cos\theta = \frac{u\cdot v}{|u||v|}[/tex]
[tex]\theta = cos^{-1}(\frac{u\cdot v}{|u||v|})[/tex]
[tex]\theta = cos^{-1}(\frac{-80}{(20\sqrt{5})(\sqrt{85})})[/tex]
[tex]\theta = cos^{-1} (-0.19402)[/tex]
[tex]\theta = 101.1868\°[/tex]
Therefore the angle would be 101.18°
