Using vectors, derive 1^° cos a=cos bcos c+sin bsin ccosα…

Using vectors, derive $1^\circ\ \ \cos a=\cos b\cos c+\sin b\sin c\cos\alpha,\qquad 2^\circ\ \ \frac{\sin\alpha}{\sin a}=\frac{\sin\beta}{\sin b}=\frac{\sin\gamma}{\sin c}$ where $a,b,c$ are the sides of a spherical triangle and $\alpha,\beta,\gamma$ the opposite angles.

Step 1: Setup

On the unit sphere, $\vec a,\vec b,\vec c$ are the unit position vectors of $A,B,C$. The sides are central angles: \[\vec b\cdot\vec c=\cos a,\qquad \vec a\cdot\vec c=\cos b,\qquad \vec a\cdot\vec b=\cos c.\]

The angle $\alpha$ at $A$ is the angle between the planes $OAB$ and $OAC$, whose normals are $\vec a\times\vec b$ and $\vec a\times\vec c$, with $|\vec a\times\vec b|=\sin c$ and $|\vec a\times\vec c|=\sin b$.

Step 2: $1^\circ$ Law of cosines

By Problem 140: \[(\vec a\times\vec b)\cdot(\vec a\times\vec c)=(\vec a\cdot\vec a)(\vec b\cdot\vec c)-(\vec a\cdot\vec c)(\vec b\cdot\vec a)=\cos a-\cos b\cos c.\]

\[ \cos\alpha=\frac{\cos a-\cos b\cos c}{\sin b\sin c}. \]

\[\cos a=\cos b\cos c+\sin b\sin c\cos\alpha\]

Cyclic permutation gives the formulas for $\cos b$ and $\cos c$.

Step 3: $2^\circ$ Law of sines

With $(\vec u\times\vec v)\times\vec w=\vec v(\vec u\cdot\vec w)-\vec u(\vec v\cdot\vec w)$ and $\vec a\cdot(\vec a\times\vec c)=0$: \[(\vec a\times\vec b)\times(\vec a\times\vec c)=-\vec a\,\big[\vec b\cdot(\vec a\times\vec c)\big]=-\vec a\,[\vec b,\vec a,\vec c]=[\vec a,\vec b,\vec c]\,\vec a.\]

Taking lengths, with $|\vec a|=1$: \[|[\vec a,\vec b,\vec c]|=|\vec a\times\vec b|\,|\vec a\times\vec c|\sin\alpha=\sin c\,\sin b\,\sin\alpha.\]

The same at $B$ and $C$ gives

\[ |[\vec a,\vec b,\vec c]|=\sin a\sin c\sin\beta=\sin a\sin b\sin\gamma. \]

Dividing by $\sin a\sin b\sin c$: \[\frac{\sin\alpha}{\sin a}=\frac{\sin\beta}{\sin b}=\frac{\sin\gamma}{\sin c}=\frac{|[\vec a,\vec b,\vec c]|}{\sin a\sin b\sin c}\]