if two vectiors of magnitudeof 5 and 3 are added such that angle between resultant and vector of magnitude 3 is maximum and it will be.

Let the two vectors $\underset{A}{\to } and \underset{B}{\to }$ inclined at an angle $\theta$ be acting on a particle at the same time.
Where, $\underset{OP}{\to } =$ $\underset{A}{\to } = 3 units$
$\underset{OS}{\to } = \underset{B}{\to } = 5 units$


According to parallelogram law of vectors, their resultant vector $\underset{R}{\to }$ wil be represented by the diagonal $\underset{OQ}{\to }$ of the parallelogram.
Let the resultant $\underset{R}{\to }$ make an angle β with the direction of $\underset{A}{\to }$. Then from right angle triangle QNO
$\tan \beta = \frac{QN}{ON} = \frac{QN}{OP + PN} = \frac{B \sin \theta }{A + B \cos \theta } .......\left(1\right)$
Angle between resultant and vector of magnitude 3 will be  maximum when $\theta = {90}^0$
From eq. 1, we get  :
$$\begin{gathered} \tan \beta = \frac{5}{3} \\ \beta = {\tan }^{-1}\left(\frac{5}{3}\right) \end{gathered}$$