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Question
If a, b, c are the lengths of sides, BC, CA and AB of a triangle ABC, prove that \[\vec{BC} + \vec{CA} + \vec{AB} = \vec{0}\] and deduce that \[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} .\]
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Solution
\[\text{ We have } \]
\[ \vec{BC} = \vec{a} \]
\[ \vec{CA} = \vec{b} \]
\[ \vec{AB} = \vec{c} \]
\[\left| \vec{a} \right|=a\]
\[\left| \vec{b} \right| =b ( \because \text{ Length is always positive} )\]
\[ \vec{c} =c \]
\[ \text{ Now } , \]
\[ \vec{BC} + \vec{CA} + \vec{AB} = \vec{0} ( \text{ Given } )\]
\[ \Rightarrow \vec{a} + \vec{b} + \vec{c} = \vec{0} \]
\[ \Rightarrow \vec{a} \times \left( \vec{a} + \vec{b} + \vec{c} \right) = \vec{a} \times \vec{0} \]
\[ \Rightarrow \vec{a} \times \vec{a} + \vec{a} \times \vec{b} + \vec{a} \times \vec{c} = \vec{0} \]
\[ \Rightarrow \vec{0} + \vec{a} \times \vec{b} - \vec{c} \times \vec{a} = \vec{0} \]
\[ \Rightarrow \vec{a} \times \vec{b} = \vec{c} \times \vec{a} \]
\[ \Rightarrow \left| \vec{a} \right| \left| \vec{b} \right|\sin C = \left| \vec{c} \right| \left| \vec{a} \right| \sin B\]
\[ \Rightarrow ab \sin C = ca \sin B\]
\[\text{ Dividing both sides by abc,we get } \]
\[ \Rightarrow \frac{\sin C}{c} = \frac{\sin B}{b} . . . (1)\]
\[\text{ Again } ,\]
\[ \vec{BC} + \vec{CA} + \vec{AB} = \vec{0} \]
\[ \Rightarrow \vec{a} + \vec{b} + \vec{c} = \vec{0} \]
\[ \Rightarrow \vec{b} \times \left( \vec{a} + \vec{b} + \vec{c} \right) = \vec{b} \times \vec{0} \]
\[ \Rightarrow \vec{b} \times \vec{a} + \vec{b} \times \vec{b} + \vec{b} \times \vec{c} = \vec{0} \]
\[ \Rightarrow - \vec{a} \times \vec{b} + \vec{0} + \vec{b} \times \vec{c} = \vec{0} \]
\[ \Rightarrow \vec{a} \times \vec{b} = \vec{b} \times \vec{c} \]
\[ \Rightarrow \left| \vec{a} \right| \left| \vec{b} \right| \sin C = \left| \vec{b} \right| \left| \vec{c} \right| \sin A\]
\[ \Rightarrow ab \sin C = bc \sin A\]
\[\text{ Dividing both sides by abc,we get } \]
\[ \Rightarrow \frac{\sin C}{c} = \frac{\sin A}{a} . . . (2)\]
\[\text{ From (1) and (2), we get } \]
\[\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}\]
\[ \Rightarrow \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\]
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