The angle $\theta$ between directions of forces, $\overrightarrow{A}$ and $\overrightarrow{B}$ is $90°$, where, $A=8$ $\mathrm{dyne}$ and $B=6$ $\mathrm{dyne}$. If, the resultant $\overrightarrow{R}$ makes an angle $\alpha$ with $\overrightarrow{A}$ then, find the value of $\alpha$(in degrees)?

The $x$ and $y$ components of the vector $\overrightarrow{A}$ are $4 \mathrm{m}$ and $6 \mathrm{m}$ respectively. The $x$, $y$ components of the vector $\overrightarrow{A}+\overrightarrow{B}$ are $10 \mathrm{m}$ and $9 \mathrm{m}$ respectively. 

If $\left|\overrightarrow{B}\right|=B \mathrm{m}$ and the angle made by $\overrightarrow{B}$ with $x$ axis is ${\tan }^{-1}\left(\frac{1}{t}\right)$ then what is the value of $B^{2}t$?