Chapter 13 - Vector Geometry - 13.3 Dot Product and the Angle Between Two Vectors - Exercises - Page 667: 50

The angles are given below (please see the figure attached): $\omega \simeq 47.49^\circ $ $\alpha \simeq 60.95^\circ $ $\beta \simeq 71.57^\circ $

Let $O$, $A$ and $B$ denote the origin, the point $\left( {2,7} \right)$ and the point $\left( {6,3} \right)$, respectively (please see the figure attached). So, $\overrightarrow {OA} = \left( {2,7} \right)$, ${\ \ }$ $\overrightarrow {AO} = - \left( {2,7} \right)$, $||\overrightarrow {OA} || = ||\overrightarrow {AO} || = \sqrt {4 + 49} = \sqrt {53} $ $\overrightarrow {OB} = \left( {6,3} \right)$, ${\ \ }$ $||\overrightarrow {OB} || = \sqrt {36 + 9} = \sqrt {45} $ $\overrightarrow {AB} = \left( {6,3} \right) - \left( {2,7} \right) = \left( {4, - 4} \right)$, ${\ \ }$ $||\overrightarrow {AB} || = \sqrt {16 + 16} = \sqrt {32} $ Let $\omega $, $\alpha $ and $\beta $ be the angles such that $\omega $: the angle between $\overrightarrow {OA} $ and $\overrightarrow {OB} $ $\alpha $: the angle between $\overrightarrow {AO} $ and $\overrightarrow {AB} $ $\beta $: the angle between $\overrightarrow {BO} $ and $\overrightarrow {BA} $ By Eq. (1) of Theorem 2, we have $\cos \omega = \frac{{\overrightarrow {OA} \cdot\overrightarrow {OB} }}{{||\overrightarrow {OA} ||||\overrightarrow {OB} ||}} = \frac{{\left( {2,7} \right)\cdot\left( {6,3} \right)}}{{\sqrt {53} \cdot\sqrt {45} }} = \frac{{11}}{{\sqrt {265} }}$ $\omega \simeq 0.829$ ${\ \ \ }$ or ${\ \ \ }$ $\omega \simeq 47.49^\circ $ $\cos \alpha = \frac{{\overrightarrow {AO} \cdot\overrightarrow {AB} }}{{||\overrightarrow {AO} ||||\overrightarrow {AB} ||}} = \frac{{ - \left( {2,7} \right)\cdot\left( {4, - 4} \right)}}{{\sqrt {53} \cdot\sqrt {32} }} = \frac{5}{{\sqrt {106} }}$ $\alpha \simeq 1.064$ ${\ \ \ }$ or ${\ \ \ }$ $\alpha \simeq 60.95^\circ $ So, $\beta = \pi - \omega - \alpha = \pi - 0.829 - 1.064$ $\beta = 1.249$ ${\ \ \ }$ or ${\ \ \ }$ $\beta \simeq 71.57^\circ $