$\text{ The resultant of }\overrightarrow{A}\text{ and }\overrightarrow{B}\text{ makes an angle }\alpha \text{ with }\overrightarrow{A}\text{ and }\beta \text{ with }\overrightarrow{B}\text{, then }$

Given that

$\overrightarrow{A},\overrightarrow{B}\text{ are two vectors having angle }\theta \text{ between them }$

They are represented as adjacent sides of parallelogram OACB as shown.

$\text{ The resultant }\mathrm{R}\text{ is making angle }\alpha \text{ with }\overrightarrow{A}\text{ and }\beta \text{ with }\overrightarrow{B}$

$\text{ Let }\left|\overrightarrow{A}\right|=a,\left|\overrightarrow{B}\right|=b$

$\text{ We have }\tan \alpha =\frac{b\sin \theta }{a+b\cos \theta },\tan \beta =\frac{a\sin \theta }{b+a\cos \theta }$

(i) If $\alpha >\beta$

$\tan \alpha >\tan \beta$

$\frac{b\sin \theta }{a+b\cos \theta }>\frac{a\sin \theta }{b+a\cos \theta }$

$\begin{array}{r}\Rightarrow b(b+a\cos \theta )>a(a+b\cos \theta )\\ \Rightarrow b^2+ab\cos \theta >a^2+ab\cos \theta \end{array}$

$\Rightarrow b>a$

$\text{ So if }\alpha >\beta \text{ We get }\mathrm{b}>\mathrm{a}\text{ or }\mathrm{a}<\mathrm{b}$

Similarly we can workout

$\text{ (ii) if }\alpha <\beta \text{ We get }\mathrm{b}<\mathrm{a}\text{ or }\mathrm{a}>\mathrm{b}$

$\text{ (iii) if }\alpha =\beta \text{ We get }\mathrm{b}=\mathrm{a}$