prove it by vector methods prove that a/sina= b/sinb = c/sinc by vector method and also proff cosine formula by it

prove that b= c cos a+ a cosc by vector method

 Let $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ be the three sides of the traingle, and let A, B, C be the angles opposite to these sides, respectively. Draw $BD\perp AC$.


(1) We know that area of $△ABC$ is $=\frac{1}{2}\left|\overrightarrow{a}\times \overrightarrow{b}\right|=\frac{1}{2}\left|\overrightarrow{b}\times \overrightarrow{c}\right|=\frac{1}{2}\left|\overrightarrow{c}\times \overrightarrow{a}\right|$

$$\begin{gathered} \Rightarrow ab\sin C=bc\sin A=ac\sin C \\ \Rightarrow \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \end{gathered}$$

Hence Proved.


(2)  In $△ABD,$$h^2=c^2-x^2$
       In $△BCD, h^2=a^2-{\left(b-x\right)}^2$

So, $$\begin{gathered} a^2-{\left(b-x\right)}^2=c^2-x^2 \\ \Rightarrow a^2-b^2+2bx-x^2=c^2-x^2 \\ \Rightarrow a^2=b^2+c^2-2bx \end{gathered}$$

Now, in $△ABD, \cos A=\frac{x}{c}\Rightarrow x=c \cos A$

Using this value of x in the equation we obtained above:-

$a^2=b^2+c^2-2bc \cos A$

This is the Cosine Formula. Hence Proved.


(3)  In $△ABD, \cos A=\frac{x}{c}$

       In $△BCD, \cos C=\frac{b-x}{a}$

So, RHS$$\begin{gathered} =c \cos A + a \cos C \\ =c\left(\frac{x}{c}\right) + a\left(\frac{b-x}{a}\right) \\ =x+b-x \\ =b=LHS \end{gathered}$$

​Hence Proved.