Let $\mathbf{\theta }$ be the angle that the vector $\overrightarrow{\mathbf{A}}$makes with the +x-axis, measured counterclockwise from that axis. Find angle $\mathbf{\theta }$ for a vector that has these components: (a)${\mathbf{A}}_{\mathbf{x}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{,}\mathbf{}{\mathbf{A}}_{\mathbf{y}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}$ ; (b)${\mathbf{A}}_{\mathbf{x}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{,}{\mathbf{A}}_{\mathbf{y}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}$ ; (c)${\mathbf{A}}_{\mathbf{x}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{,}{\mathbf{A}}_{\mathbf{y}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}$ ; (d) ${\mathbf{A}}_{\mathbf{x}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{,}{\mathbf{A}}_{\mathbf{y}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}$.

Here different coordinates of the vector are given. We are knowing that in coordinates x component of the vector will be first and then the y component. Now to find the angle of the vector with respect to the x-axis we just need to take the ratio of the y component to that of the x component and then take the tan inverse of that value.

This will give you the angle of the vector with the x-axis.

It can be represented as

$\mathbf{\theta }\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{y}{x}\right)$…………………(1)

Substitute the given data in equation (1), and we get,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right) \\ ={\tan }^{-1}\left(\frac{-1.00m}{2.00m}\right) \\ ={\tan }^{-1}\left(\frac{-1}{2}\right) \\ =-26.6° \\ =360°-26.6° \\ =333.4° \end{gathered}$$

Substitute the given data in equation (1), and we get,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right) \\ ={\tan }^{-1}\left(\frac{1.00m}{2.00m}\right) \\ ={\tan }^{-1}\left(\frac{1}{2}\right) \\ =26.6° \end{gathered}$$

Substitute the given data in equation (1), and we get,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right) \\ ={\tan }^{-1}\left(\frac{-1.00m}{2.00m}\right) \\ ={\tan }^{-1}\left(\frac{-1}{2}\right) \\ =-26.6° \\ =360°-26.6° \\ =333.4° \end{gathered}$$

Substitute the given data in equation (1), and we get,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right) \\ ={\tan }^{-1}\left(\frac{-1.00m}{-2.00m}\right) \\ ={\tan }^{-1}\left(\frac{1}{2}\right) \\ =26.6° \end{gathered}$$

Therefore the angle made by the vectors is (a)$333.4°$, (b)$26.6°$, (c)$333.4°$, and

(d)$26.6°$.