$A$ force of $5 \, N$ acts at an angle of $60^\circ$ with the vertical. Find the vertical component of the force in $N$.

The velocity of a particle having a magnitude of $10 \ m/s$ in the direction of $60^{\circ}$ with the positive $X$-axis is:

The angle made by the vector $\vec{A} = \hat{i} + \hat{j}$ with the $x$-axis is ....... $^\circ$.

If $\vec{A} = \hat{i} A \cos \theta + \hat{j} A \sin \theta$ is a vector,then another vector $\vec{B}$ which is perpendicular to $\vec{A}$ is given by:

Two forces of magnitude $P$ and $Q$ acting at a point have a resultant $R$. The resolved part of $R$ in the direction of $P$ is of magnitude $Q$. The angle between the forces is:

$A$ person pushes a box kept on a horizontal surface with a force of $100\,N$ as shown in the figure. In unit vector notation,the force $\vec{F}$ can be expressed as?