Question

If $a,b,c$ are the lengths of sides $BC,CA$ and $AB$ of a triangle $ABC$, prove that $\overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{0}$ and deduce that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

Answer 1

As $A,B,C$ are vertices of triangle so we know that, $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$

$\therefore \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=(\overrightarrow{AB}+\overrightarrow{BC})+\overrightarrow{CA}=\overrightarrow{AC}+\overrightarrow{CA}=0$

Now In Triangle $ABC$, The area of triangle $ABC=\frac{1}{2}|AB\times AC|=\frac{1}{2}|AB\times BC|=\frac{1}{2}|BC\times AC|$

$=\frac{1}{2}|AB||AC|sinA$ $=\frac{1}{2}|CB||AB|sinB$ $=\frac{1}{2}|BC||CA|sinC$

$\frac{b.c.sin\left(A\right)}{2}=\frac{a.c.sin\left(B\right)}{2}=\frac{a.b.sin\left(C\right)}{2}$

$b.c.sin\left(A\right)=a.c.sin\left(B\right)=a.b.sin\left(C\right)$

Divide by $abc$, we get,

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

$\Rightarrow \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$