Vectors Algebra – Sine Rule
Sine Rule: If A, B and C are the vertices of a triangle ABC, then sine rule \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\),
Let \(\overrightarrow{BC}=\overrightarrow{a}\),
\(\overrightarrow{CA}=\overrightarrow{b}\),
\(\overrightarrow{AB}=\overrightarrow{c}\),
So that \(\overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}\),
Therefore \(\overrightarrow{a}\times \overrightarrow{a}+\overrightarrow{a}\times \overrightarrow{b}=-\overrightarrow{a}\times \overrightarrow{c}\),
\(\left( \because
\overrightarrow{a}\times \overrightarrow{a}=0\ ,-\overrightarrow{a}\times
\overrightarrow{c}\ =\overrightarrow{c}\times \overrightarrow{a}\ \right)\),
\(0+\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}\times \overrightarrow{a}\),
\(\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}\times \overrightarrow{a}\),
\(\left| \overrightarrow{a}\times \overrightarrow{b} \right|=\left| \overrightarrow{c}\times \overrightarrow{a} \right|\),
\(\left( \because|\overrightarrow{a}\times \overrightarrow{b}|=a.b\sin \theta \ \right)\),
\(a.b\ \sin ({{180}^{o}}-C)=c.a\sin ({{180}^{o}}-B)\),
\(\left( \because \sin ({{180}^{o}}-\theta )\ =\sin \theta \right)\),
\(a.b\ \sin (C)=c.a\sin (B)\),
Dividing both sides by abc, we get
\(\frac{a.b\ \sin C}{abc}=\frac{c.a\sin B}{abc}\),
\(\frac{\ \sin C}{c}=\frac{\sin B}{b}\) . . .(1)
\(\frac{b}{\sin B}=\frac{c}{\sin C}\), . . .(1)
Similarly, . . .(2) \(\frac{c}{\sin C}=\frac{a}{\sin A}\), . . .(2)
From (1) and (2), we have
\(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\),
A, B and C are the vertices of a triangle ABC, then sine rule \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\).