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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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In a triangle ABC, prove that $\Large\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

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  • In a $\Delta\:ABC,\:\:\overrightarrow {AB}+\overrightarrow {BC}+\overrightarrow {CA}=\overrightarrow 0$
  • $\overrightarrow a\times\overrightarrow b=|\overrightarrow a||\overrightarrow b|sin\theta=ab\:sin\theta $ where $\theta$ is the angle between the vectors.
  • $\overrightarrow a\times\overrightarrow a=\overrightarrow 0$
In a $\Delta \:ABC$ let $\overrightarrow {AB}=\overrightarrow c$, $\overrightarrow {BC}=\overrightarrow a$ and $\overrightarrow {CA}=\overrightarrow b$
$|\overrightarrow a|= a=$side opposite to angle A, $|\overrightarrow b|= b=$side opposite to angle B
and $|\overrightarrow c|=c=$side opposite to angle C.
We know that in a $\Delta\:ABC,\:\:\overrightarrow {AB}+\overrightarrow {BC}+\overrightarrow {CA}=\overrightarrow 0$
$\Rightarrow\: \overrightarrow c+\overrightarrow a+\overrightarrow b=\overrightarrow 0$
$\Rightarrow\:\overrightarrow a+\overrightarrow b=-\overrightarrow c$
Taking cross product with $\overrightarrow c$ on both the sides we get
$(\overrightarrow a+\overrightarrow b)\times\overrightarrow c=-\overrightarrow c\times\overrightarrow c$
But we know that $\overrightarrow c\times\overrightarrow c=\overrightarrow 0$
$\Rightarrow \overrightarrow a\times\overrightarrow c+\overrightarrow b\times\overrightarrow c=\overrightarrow 0$
$\Rightarrow|\overrightarrow a||\overrightarrow c|sin(\pi+B)+|\overrightarrow b||\overrightarrow c|sin(\pi-A)=0$
$\Rightarrow \:\:-ac\:sinB+bc\:sinA=0$
$\Rightarrow \:\large\frac{a}{sinA}=\frac{b}{sinB}$.............(i)
Similarly by writing $\overrightarrow c+\overrightarrow a+\overrightarrow b=\overrightarrow 0$ as
$\Rightarrow\:\overrightarrow b+\overrightarrow c=-\overrightarrow a$ and
taking cross product with $\overrightarrow a$ on both the sides we get
$(\overrightarrow b+\overrightarrow c)\times\overrightarrow a=-\overrightarrow a\times\overrightarrow a$
But we know that $\overrightarrow a\times\overrightarrow a=\overrightarrow 0$
$\Rightarrow \overrightarrow b\times\overrightarrow a+\overrightarrow c\times\overrightarrow a=\overrightarrow 0$
$\Rightarrow|\overrightarrow b||\overrightarrow a|sin(\pi+C)+|\overrightarrow c||\overrightarrow a|sin(\pi-B)=0$
$\Rightarrow \:\:-ba\:sinC+ca\:sinB=0$
$\Rightarrow \:\large\frac{c}{sinC}=\frac{b}{sinB}$.............(ii)
From (i) and (ii) we get the result
Hence proved.
answered Apr 20, 2013 by rvidyagovindarajan_1

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