In the sum $\overrightarrow{\mathbf{A}}\mathbf{+}\overrightarrow{\mathbf{B}}\mathbf{+}\overrightarrow{\mathbf{C}}$vector has a magnitude of 12.0 m and is angled $\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{°}$counterclockwise from the +x direction, and vector $\overrightarrow{\mathbf{C}}$has a magnitude of 15.0 m and is angled $\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{°}$counterclockwise from the -x direction. What are (a) the magnitude and (b) the angle (relative to +x ) of $\overrightarrow{\mathbf{B}}$?

Resolving vectors into their components using the vector algebra unknown vector can be found. The magnitude and the angle between the vector and +x axis can be calculated using the general equations given below for the magnitude and the angle.

Formulae

$$\begin{gathered} \overrightarrow{A}=\overrightarrow{A_x}+\overrightarrow{A_y}=A\cos \theta +A\sin \theta \\ \overrightarrow{C}=\overrightarrow{C_x}+\overrightarrow{C_y}=C\cos \theta +C\sin \theta \\ \overrightarrow{A}=\overrightarrow{B}+\overrightarrow{C} \\ {B}_x=C_x+A_x=C\cos \theta -A\cos \theta \\ {B}_y=C_y+A_y=C\sin \theta -A\sin \theta \\ {B}_y=\left|\overrightarrow{B}\right|=\sqrt{{\left(\overrightarrow{B_x}\right)}^2+{\left(\overrightarrow{B_y}\right)}^2} \\ \theta ={\tan }^{-1}\left(\frac{\overrightarrow{B_y}}{B_x}\right) \\ \mathrm{Given}\mathrm{are}, \\ \mathrm{The}\mathrm{magnitude}\mathrm{of}\mathrm{vector}\overrightarrow{\mathrm{A}}=12\mathrm{m}\mathrm{and}\mathrm{angle}\mathrm{between}\overrightarrow{\mathrm{A}}\mathrm{and}+\mathrm{x}\mathrm{axis}\mathrm{is}40° \\ \mathrm{The}\mathrm{magnitude}\mathrm{of}\mathrm{vector}\overrightarrow{\mathrm{C}}=15\mathrm{m}\mathrm{and}\mathrm{angle}\mathrm{between}\overrightarrow{\mathrm{C}}\mathrm{and}+\mathrm{x}\mathrm{axis}\mathrm{is} \\ 180°+20°=200° \end{gathered}$$

$$\begin{gathered} {\overrightarrow{B}}_x={\overrightarrow{C}}_x-{\overrightarrow{A}}_x=\left(15.0\mathrm{m}\right)\cos \left(200\right)-\left(12.0\mathrm{m}\right)\cos \left(40\right) \\ {\overrightarrow{\mathrm{B}}}_{\mathrm{x}}=-23.3\mathrm{m} \\ {\overrightarrow{\mathrm{B}}}_{\mathrm{y}}={\overrightarrow{\mathrm{C}}}_{\mathrm{x}}-{\overrightarrow{\mathrm{A}}}_{\mathrm{x}}=\left(15.0\mathrm{m}\right)\sin \left(200\right)-\left(12.0\mathrm{m}\right)\sin \left(40\right) \\ {\overrightarrow{\mathrm{B}}}_{\mathrm{y}}=-12.8\mathrm{m} \end{gathered}$$

The magnitude of vector $\overrightarrow{B}$is

$\sqrt{{\left(-23.3\mathrm{m}\right)}^2+{\left(-12.8\mathrm{m}\right)}^2}=26.6\mathrm{m}$

The angle of vector $\overrightarrow{B}$ relative to +x axis is

${\tan }^{-1}\left(\frac{-12.8}{-23.3}\right)=28.9°$

As vector $\overrightarrow{B}$is in third quadrant,

$180°+28.9°=209°$