Example 1.22. Monica and Linda traveled from Shanghai, China (east $121^{\circ}$, north $31^{\circ}$ ), to Albany, New York (west $73^{\circ}$, north $42^{\circ}$ ), to visit their friend Hilary. Monica's hometown, Billrock, is four-fifths of the way along their trip. Find the spherical coordinates of the town of Billrock. Figure 1.57. 58 103 Trigonometry Problems Solution: We maintain the same notation as in the solution of Example 1.21. Let $C$ denote an arbitrary point on $\widehat{A B}$, and assume that $\angle A O C=k \angle A O B=k \theta$, where is $k$ is some real number with $0 \leq k \leq 1$. (See Figure 1.57.) Then $\angle C O B=(1-k) \theta$. Let $D$ and $E$ be points on lines $O A$ and $O B$, respectively, such that $A O \| C E$ and $B O \| C D$. Then $C D O E$ is parallelogram. Set $\mathbf{u}=\overrightarrow{O A}, \mathbf{v}=\overrightarrow{O B}$, and $\mathbf{w}=\overrightarrow{O C}$. There are real numbers $a$ and $b$ such that $|O D|=a \cdot|O A|$ and $|O E|=b \cdot|O B|$. Then $\mathbf{w}=a \mathbf{u}+b \mathbf{v}$. By the distributive property of the dot product, we have $$ \mathbf{w} \cdot \mathbf{u}=a \mathbf{u} \cdot \mathbf{u}+b \mathbf{u} \cdot \mathbf{v} \text { and } \mathbf{w} \cdot \mathbf{v}=a \mathbf{u} \cdot \mathbf{v}+b \mathbf{v} \cdot \mathbf{v} . $$ Note that $|\mathbf{u}|=|\mathbf{v}|=|\mathbf{w}|=3960$. By the vector form of the law of cosines, these equations are equivalent to the equations $$ \cos k \theta=a+b \cos \theta \text { and } \cos (1-k) \theta=a \cos \theta+b . $$ Solving this system of equations for $a$ and $b$ gives $$ a=\frac{\cos k \theta-\cos (1-k) \theta \cos \theta}{\sin ^2 \theta} \text { and } b=\frac{\cos (1-k) \theta-\cos k \theta \cos \theta}{\sin ^2 \theta} . $$ For our problem, we have $k=\frac{4}{5}$ and $\theta \approx 105.870^{\circ}$ (by Example 1.21). Substituting these values into the above equations gives $a \approx 0.376$ and $b \approx 1.035$. It follows that $\mathbf{w}=a \mathbf{u}+b \mathbf{v} \approx 3960[0.059,-0.460,0.886]$, which are the rectangular coordinates of the point $C$. Let $(3960, \gamma, \phi)$ be the spherical coordinates of $C$. Then $\sin \phi \approx$ 0.886 , or $\phi \approx 62.383^{\circ}$, and $\sin \gamma \cos \phi \approx-0.460$, or $\gamma \approx-82.67^{\circ}$; that is, Billrock has longitude west $82.67^{\circ}$ and latitude $62.383^{\circ}$.
This is a problem taken from the 103 Trigonometric Problems book. specifically, it introduces vectors in relation to angles, my particular question with regard to this solution is the identities. I do not seem to understand what Vector form of the law of cosines that they used to get the inequalities $$ \cos k \theta=a+b \cos \theta \quad $$ and $$ \quad \cos (1-k) \theta=a \cos \theta+b $$.
Out of context, I would have assumed that the vector form of the law of cosines was the formula shown in this answer, $\|\mathbb v_1 + \mathbb v_2\|^2 = \|\mathbb v_1\|^2+\|\mathbb v_2\|^2+2 \|\mathbb v_1\| \|\mathbb v_2\| \cos\theta$. But in context, it appears that this author is referring only to one of the steps in the derivation of the last term of that identity, that is, the fact that
$$ \mathbb v_1 \cdot \mathbb v_2 = \|\mathbb v_1\| \|\mathbb v_2\| \cos\theta, \tag1 $$
where $\theta$ is the angle between the vectors $V_1$ and $V_2$.
Quite likely this meaning of "the vector form of the law of cosines" has been introduced somewhere in the previous 57 pages of the book.
What I believe the author expects you to do is to apply Equation $(1)$ to your two formulas, $\mathbf{w} \cdot \mathbf{u} = a \mathbf{u} \cdot \mathbf{u}+b \mathbf{u} \cdot \mathbf{v}$ and $\mathbf{w} \cdot \mathbf{v} = a \mathbf{u} \cdot \mathbf{v}+b \mathbf{v} \cdot \mathbf{v}$, in order to replace the dot products with ordinary products of $\|\mathbf u\|$, $\|\mathbf v\|$, $\|\mathbf w\|$, and the cosines of angles between those vectors. You can refer to the angles in the figure that came with the exercise:
For dot products like $\mathbf u \cdot \mathbf u$, of course, the angle is zero and the product just comes out to $\|\mathbf u\|^2$.
After you have finished rewriting all the dot products this way, look for common factors, remembering that $\|\mathbf u\| = \|\mathbf v\| = \|\mathbf w\|$. By dividing all terms by the common factor you should end up with $\cos k \theta=a+b \cos \theta$ and $\cos (1-k) \theta=a \cos \theta+b$.