Question

Using vector method prove that in a ∆ABC a/sinA = b/sinB = c/SinC

where , a ,b , c are the length of the sides opposite respectively to the angles A, B , And C of ∆ABC

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Important question for 12th boards .​

Answer 1

Step-by-step explanation:

Solution is in this given attachment

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Answer 2

The image of the triangle is attached below.

Let $\overrightarrow{A B}=\bar{c}, \overrightarrow{B C}=\bar{a}, \overrightarrow{C A}=\bar{b}$

$\bar{a}+\bar{b}+\bar{c}=\overline{0}$

By the definition of cross Product,

$\bar{a} \times(\bar{a}+\bar{b}+\bar{c})=\bar{a} \times \overline{0}$

By the property of distributive $\bar{x} \times(\bar{y}+\bar{x})=\bar{x} \times \bar{y}+\bar{x} \times \bar{z}$

$\Rightarrow(\bar{a} \times \bar{a})+(\bar{a} \times \bar{b})+(\bar{a} \times \bar{c})=\overline{0} \rightarrow[\because \bar{a} \times \overline{0}=\overline{0}]$

We know that $\bar{a} \times \bar{a}=\overline{0} \Rightarrow \hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=\overline{0}$

$\Rightarrow \overline{0}+(\bar{a} \times \bar{b})+(\bar{a} \times \bar{c})=\overline{0} \rightarrow[\because(\bar{a} \times \bar{a})=\overline{0}]$

By the property of non-commutative $\bar{x} \times \bar{y}=-(\bar{y} \times \bar{x})$

$\Rightarrow(\bar{a} \times \bar{b})-(\bar{c} \times \bar{a})=\overline{0}$

$\Rightarrow(\bar{a} \times \bar{b})=(\bar{c} \times \bar{a})$

$\Rightarrow a \cdot b \sin (\pi-C)=c \cdot a \sin (\pi-B)$

Since $\sin\pi=0$

$\Rightarrow b \sin C=c \sin B$

$\Rightarrow \frac{\sin C}{c}=\frac{\sin B}{b}$ – – – – (1)

Similarly we can prove that ,

$\Rightarrow \frac{\sin A}{a}=\frac{\sin B}{b}$ – – – – (2)

Hence,

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