91Ó°ÊÓ

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

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

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

The magnitude of vector $\stackrel{→}{B}$is

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

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

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

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

$180°+28.9°=209°$