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Applications of Dot Product

In any triang...

Question

In any triangle ABC, prove by vector method
$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$

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Solution

In $Δ A B C$ , Let $- - \to B C = \to a, - - \to C A = \to b$ and $- - \to A B = \to c$ , then
$a = | - - \to B C |, b = | - - \to C A |$ and $c = | - - \to A B |$
From $Δ A B C$ , we get
$\Rightarrow - - \to B C + - - \to C A = - - \to B A$
$\Rightarrow - - \to B C + - - \to C A + - - \to A B = 0$
$\Rightarrow \to a + \to b + \to c = 0$
$\Rightarrow \to a \times (\to a + \to b + \to c) = \to a \times \to 0$
$\Rightarrow \to a \times \to a + \to a \times \to b + \to a \times \to c = 0$
$\Rightarrow \to a \times \to b = - \to a \times \to c$
$\Rightarrow \to a \times \to b = \to c \times \to a$ ...... (i)
Similarly, $\to a \times \to b = \to b \times \to c$ ...... (ii)
From (i) and (ii), we get
$\to a \times \to b = \to b \times \to c = \to c \times \to a$
$| \to a \times \to b | = | \to b \times \to c | = | \to c \times \to a |$
$a b sin (π - C) = b c sin (π - A) = c a sin (π - B)$
$a b sin C = b c sin A = c a sin B$
Dividing by abc throughout, we get
$\frac{sin C}{c} = \frac{sin A}{a} = \frac{sin B}{b}$
$\Rightarrow \frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$

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