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

Given that

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

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

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

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

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

(i) If $\alpha >\beta$

$\tan \alpha >\tan \beta$

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

$\Rightarrow b(b+a\cos \theta )>a(a+b\cos \theta )$

$\Rightarrow b^2+ab\cos \theta >a^2+ab\cos \theta$

$\Rightarrow b>a$

So  if  $\alpha >\beta$ We get  b > a or a < b

Similarly we can workout

(ii) if  $\alpha <\beta$We get  b < a or a > b

(iii) if $\alpha =\beta$ We get  b = a