Using vectors, prove that in a ABC, a = b = c Where a, b and c ar…

Question

Using vectors, prove that in a $\Delta$ ABC,
$\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}$
Where a, b and c are lengths of the sides opposite, respectively, to the angles A, B and C of $\Delta$ ABC.

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Answer

Let in $\Delta$ ABC, BC = $\vec{\text{a}},\text{ }\text{ CA}=\vec{\text{b}}\text{ and }\text{ }\vec{\text{AB}}=\vec{\text{c}}$
$\therefore\text{ }\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\vec{\text{o}}$
$\vec{\text{a}}\times\vec{\text{a}}+\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{a}}\times\vec{\text{c}}=\vec{\text{o}}\Rightarrow\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{c}}\times\vec{\text{a}}............\text{(i)}$
and $\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{b}}+\vec{\text{c}}\times\vec{\text{b}}=\vec{\text{o}}\Rightarrow\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{b}}\times\vec{\text{c}}..........{\text{(ii)}}$
$\therefore\text{ }\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{b}}\times\vec{\text{c}}=\vec{\text{c}}\times\vec{\text{a}}\Rightarrow\Bigg[\vec{\text{a}}\times\vec{\text{b}}\Bigg]=\Bigg[\vec{\text{b}}\times\vec{\text{c}}\Bigg]=\Bigg[\vec{\text{c}}\times\vec{\text{a}}\Bigg]$
$\therefore\text{ab sin C = bc sin A = ca sin B }\text{ Or }\frac{\sin\text{C}}{\text{c}}=\frac{\sin\text{A}}{\text{a}}=\frac{\sin\text{B}}{\text{b}}$
$\therefore\frac{\text{a}}{\text{sin A}}=\frac{\text{b}}{\text{sin B}}=\frac{\text{c}}{\text{sin C}}.$

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