Question

Use vectors to prove a/SinA= b/SinB = c/SinC for ΔABC.

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Let us consider the triangle given below.

$\overrightarrow{B C}=\vec{a} ; \overrightarrow{C A}=\vec{b} ; \overrightarrow{A B}=\vec{c}$

From triangle,

$\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$

Consider:

$\vec{a} \times(\vec{a}+\vec{b}+\vec{c})=\vec{a} \times \overrightarrow{0}$

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

$0+(\vec{a} \times \vec{b})-(\vec{c} \times \vec{a})=0$

$(\vec{a} \times \vec{b})=(\vec{c} \times \vec{a}) \rightarrow(1)$

Similarly,

$\vec{b} \times(\vec{a}+\vec{b}+\vec{c})=\vec{b} \times \overrightarrow{0}$

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

$-(\vec{a} \times \vec{b})+(\vec{b} \times \vec{b})+(\vec{b} \times \vec{c})=0$

$-(\vec{a} \times \vec{b})+0+(\vec{b} \times \vec{c})=0$

$(\vec{a} \times \vec{b})=(\vec{b} \times \vec{c}) \rightarrow(2)$

From (1) and (2), we get,

$(\vec{a} \times \vec{b})=(\vec{b} \times \vec{c})=(\vec{c} \times \vec{a})$

$|\vec{a}||\vec{b}| \sin (\pi-C)=|\vec{b}||\vec{c}| \sin (\pi-A)=|\vec{c}||\vec{a}| \sin (\pi-B)$

$a b \sin C=b c \sin A=c a \sin B$

On dividing abc, we get,

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

Hence proved.