Question
Using vector method prove that in any triangle ABC,
$
\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} .
$
$
\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} .
$
✓
Answer
Let in $\triangle ABC$
$\overrightarrow{ BC }=\vec{a}, \overrightarrow{ CA }=\vec{b}$ and $\overrightarrow{ AB }=\vec{c}$
$\therefore \quad|\overrightarrow{ BC }|=a,|\overrightarrow{ CA }|=b$ and $|\overrightarrow{ AB }|=c$
Now in $\triangle ABC$ by triangle law of vector addition,
$\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{BA}
$
or $
\overrightarrow{BC}+\overrightarrow{CA}=-\overrightarrow{AB}
$
or $
\overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AB}=0
$
or $
\vec{a}+\vec{b}+\vec{c}=0
$
Now
$
\vec{a} \times(\vec{a}+\vec{b}+\vec{c})=\vec{a} \times 0
$
or $\quad(\vec{a} \times \vec{a})+(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})=0 \quad$ [by distribution law]
$
\begin{aligned}
or \quad \vec{a} \times \vec{b}+\vec{a} \times \vec{c} & =0 \\
\end{aligned}
$
$
\begin{array}{ll}
\text { or } & \vec{a} \times \vec{b}=-\vec{a} \times \vec{c} \\
\text { or } & \vec{a} \times \vec{b}=\vec{c} \times \vec{a}
\end{array}
$
$
|\vec{a} \times \vec{b}|=|\vec{c} \times \vec{a}|
$
Similarly from $\vec{b} \times(\vec{a}+\vec{b}+\vec{c})=\overrightarrow{0}$ we can find that
$
|\vec{b} \times \vec{c}|=|\vec{a} \times \vec{b}|
$
Hence from equations (2) and (3),
$
|\vec{b} \times \vec{c}|=|\vec{c} \times \vec{a}|=|\vec{a} \times \vec{b}|
$
or $b c \sin (\pi- A )=c a \sin (\pi- B )=a b \sin (\pi- C )$
or $\quad b c \sin A=c a \sin B=a b \sin C$
Dividing $a b c$ in equation ....... (4),
$
\frac{b c \sin A}{a b c}=\frac{c a \sin B}{a b c}=\frac{a b \sin C}{a b c}
$
or $
\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}
$
Hence proved.
$\overrightarrow{ BC }=\vec{a}, \overrightarrow{ CA }=\vec{b}$ and $\overrightarrow{ AB }=\vec{c}$
$\therefore \quad|\overrightarrow{ BC }|=a,|\overrightarrow{ CA }|=b$ and $|\overrightarrow{ AB }|=c$
Now in $\triangle ABC$ by triangle law of vector addition,
$\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{BA}
$
or $
\overrightarrow{BC}+\overrightarrow{CA}=-\overrightarrow{AB}
$
or $
\overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AB}=0
$
or $
\vec{a}+\vec{b}+\vec{c}=0
$
Now
$
\vec{a} \times(\vec{a}+\vec{b}+\vec{c})=\vec{a} \times 0
$
or $\quad(\vec{a} \times \vec{a})+(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})=0 \quad$ [by distribution law]
$
\begin{aligned}
or \quad \vec{a} \times \vec{b}+\vec{a} \times \vec{c} & =0 \\
\end{aligned}
$
$
\begin{array}{ll}
\text { or } & \vec{a} \times \vec{b}=-\vec{a} \times \vec{c} \\
\text { or } & \vec{a} \times \vec{b}=\vec{c} \times \vec{a}
\end{array}
$
$
|\vec{a} \times \vec{b}|=|\vec{c} \times \vec{a}|
$
Similarly from $\vec{b} \times(\vec{a}+\vec{b}+\vec{c})=\overrightarrow{0}$ we can find that
$
|\vec{b} \times \vec{c}|=|\vec{a} \times \vec{b}|
$
Hence from equations (2) and (3),
$
|\vec{b} \times \vec{c}|=|\vec{c} \times \vec{a}|=|\vec{a} \times \vec{b}|
$
or $b c \sin (\pi- A )=c a \sin (\pi- B )=a b \sin (\pi- C )$
or $\quad b c \sin A=c a \sin B=a b \sin C$
Dividing $a b c$ in equation ....... (4),
$
\frac{b c \sin A}{a b c}=\frac{c a \sin B}{a b c}=\frac{a b \sin C}{a b c}
$
or $
\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}
$
Hence proved.
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