The x component of vector $\overrightarrow{A}$is -25.02 m and the y component is +40.0 m. (a) What is the magnitude of $\overrightarrow{A}$? (b) What is the angle between the direction of $\overrightarrow{A}$ and the positive direction of x?

In a two-dimensional coordinate system, vectors can be split into component x and component y. In this problem, x and y components of vector $\overrightarrow{A}$are given. Using these components, the magnitude of the vector is calculated. Also, the magnitude angle notation can be used to find out the direction of the vector. Vector $\overrightarrow{A}$ can be represented in the magnitude angle notation,

$$\begin{gathered} A=\sqrt{A_x^2+A_y^2} \\ \vartheta =\left(\frac{A_y}{A_x}\right) \\ \mathrm{Given}\mathrm{parameters}, \\ {\mathrm{A}}_{\mathrm{x}}=-25.0\mathrm{m} \\ {\mathrm{A}}_{\mathrm{y}}=+40.0\mathrm{m} \\ \mathrm{substituting}\mathrm{the}\mathrm{values}\mathrm{of}{\mathrm{A}}_{\mathrm{x}}\mathrm{and}{\mathrm{A}}_{\mathrm{y}}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{the}\mathrm{magnitude}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}, \\ \mathrm{A}=\sqrt{{(-25)}^2+{\left(40\right)}^2} \\ \mathrm{A}=47.2\mathrm{m} \end{gathered}$$

substituting the values of $A_x$and $A_y$in equation (ii), the direction can be written as,

$\theta =\left(\frac{A_y}{A_x}\right)=\left(\frac{40}{-25}\right)=-{58}^{°}$

The angle between $\overrightarrow{A}$and the negativex axis is $-{58}^{°}$. By adding ${180}^{°}$, we get ${122}^{°}$ which is the angle from the positivex axis.