Precalculus (6th Edition) Blitzer

${{38.70}^{{}^\circ }}$
Let the angle between $\mathbf{v}$ and $\mathbf{w}$ be $\theta$ such that the angle between the vectors $\mathbf{v}$ and $\mathbf{w}$ can be obtained using the formula $\theta ={{\cos }^{-1}}\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left| \mathbf{v} \right|\left| \mathbf{w} \right|} \right)$ as, \begin{align} & \theta ={{\cos }^{-1}}\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left| \mathbf{v} \right|\left| \mathbf{w} \right|} \right) \\ & ={{\cos }^{-1}}\left( \frac{\left( 6\mathbf{i}+0\mathbf{j} \right)\cdot \left( 5\mathbf{i}+4\mathbf{j} \right)}{\left( \sqrt{{{\left( 6 \right)}^{2}}} \right)\left( \sqrt{{{5}^{2}}\mathbf{+}{{4}^{2}}} \right)} \right) \\ & ={{\cos }^{-1}}\left( \frac{6\cdot 5+0\cdot 4}{\left( 6 \right)\left( \sqrt{41} \right)} \right) \\ & ={{\cos }^{-1}}\left( \frac{30}{\sqrt{1476}} \right) \end{align} Solve ahead to get the result as, \begin{align} & \theta ={{\cos }^{-1}}\left( \frac{30}{\sqrt{1476}} \right) \\ & ={{38.70}^{{}^\circ }} \end{align} Hence, the angle between $\mathbf{v}$ and $\mathbf{w}$ is $\theta ={{38.70}^{{}^\circ }}$.