Question

Prove by the vector method, the law of sine in trignometry:
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

Answer 1

Let $a,b,c$ be the sides of $DABC$ then,
$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{o}......\left(1\right)$
$\overrightarrow{a}\times (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{a}\times \overrightarrow{o}$
or $\overrightarrow{a}\times \overrightarrow{a}+\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}=\overrightarrow{o}$
$\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}\times \overrightarrow{a}.....\left(2\right)$
or $\overrightarrow{b}\times \overrightarrow{a}+\overrightarrow{b}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}=0$
or $\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{b}\times \overrightarrow{c}.....\left(3\right)$
By equation (2) and (3)
$\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{c}\times \overrightarrow{a}$
$|\overrightarrow{a}\times \overrightarrow{b}|=|\overrightarrow{b}\times \overrightarrow{c}|=|\overrightarrow{c}\times \overrightarrow{a}|$
$ab\sin (p-C)=bc\sin (p-A)=ca\sin (p-B)$
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$