The two individual movement vectors are arranged tip-to-tail from a starting point in the bottom right corner. The displacement (resultant) vector is then added, from the tail of the first vector to the tip of the second vector. We label it $\stackrel{\rightharpoonup }{d}$. We can also mark the resultant's direction angle, θ (the Greek letter theta, commonly used in physics to represent unknown angles. We know the angle between the two vectors within the triangle is 110°, since the second vector is 20° north of west, and there is a 90° between west and south.

${\left|\overrightarrow{d}\right|}^2={100}^2+{80}^2-2\left(100\right)\left(80\right)\cos \left(110°\right)$

${\left|\overrightarrow{d}\right|}^2=\mathrm{21,872}.32$

$\left|\overrightarrow{d}\right|\mathrm{ }=147.9 \mathrm{km}$

$\frac{100}{\text{sin}\mathrm{\theta }}=\frac{147.9}{\text{sin}{110}^{\circ }}$

$\sin \mathrm{\theta }\mathrm{ }=\frac{100\sin 1{10}^{\circ }}{147.9}$

$\mathrm{\theta }\mathrm{ }={39.4}^{\circ }$

$\therefore \overrightarrow{d}=148 \mathrm{km} \left[\mathrm{N} \mathrm{39°} \mathrm{W}\right]$

Therefore, the displacement of the streetcar is 148 km [N 39° W].

$\left|\overrightarrow{d_x}\right|=\left|\overrightarrow{d}\right|\cos \mathrm{\theta }$

$\left|\overrightarrow{d_x}\right|=\left|5.0 \mathrm{m}\right|\cos 45^{\circ }$

$\left|\overrightarrow{d_x}\right|=3.5 \mathrm{m}$

$\left|\overrightarrow{d_y}\right|=\left|\overrightarrow{d}\right|\sin \mathrm{\theta }$

$\left|\overrightarrow{d_y}\right|=\left|5.0 \mathrm{m}\right|\sin 45^{\circ }$

$\left|\overrightarrow{d_y}\right|=3.5 \mathrm{m}$

Start by defining which directions are positive. Typically, north and east (up and right on a diagram) will be treated as positive, but we always have the option to change that. [North] and [East] are the positive directions.

For this example, the following diagram includes the perpendicular components.

Vector $\overrightarrow{d_1}$ is already vertical, so it has just one component: 80 km [N]. Vector $\overrightarrow{d_2}$ has a vertical and a horizontal component: 34.2 km [N] and 94 km [W]. In the previous step, we defined [East] as a positive direction, so we will have to be mindful of that [West] in the next step.

You can explore how the components in the previous diagram were obtained:

$\overrightarrow{d_1}=80.0 \mathrm{km} \left[N\right]$

$\overrightarrow{d_2}=1.0\times {10}^2 \mathrm{km} \left[\mathrm{W} {20}^{\circ } \mathrm{N}\right]$

$\overrightarrow{d_{1x}}=0$

$\overrightarrow{d_{1y}}=80 \mathrm{km}$

$\overrightarrow{d_{2x}}=-\left(100 \mathrm{km}\right)\cos 20^{\circ }=-94.0 \mathrm{km}$

$\overrightarrow{d_{2y}}= \left(100 \mathrm{km}\right)\sin 20^{\circ }=34.2 \mathrm{km}$

Now that we have the individual components, we can find the components of the resultant by just adding (or subtracting) parallel components.

$d_{Tx}=d_{1x}+d_{2x}=0+(-94.0 \mathrm{km})=-94.0 \mathrm{km}$

$d_{Ty}= \left|d_{1y}\right|+\left|d_{2y}\right|=80 \mathrm{km}+34.2 \mathrm{km}=114.2 \mathrm{km}$

Use Pythagorean theorem and tangent ratio to determine the magnitude of the resultant vector and its direction (angle).

$\left|\overrightarrow{d_T}\right|=\sqrt{{\left|\overrightarrow{d_x}\right|}^2+{\left|\overrightarrow{d_y}\right|}^2}$

$\left|\overrightarrow{d_T}\right|=\sqrt{{\left(-94.0 \mathrm{km}\right)}^2+{\left(114.2 \mathrm{km}\right)}^2}$

$\left|\overrightarrow{d_T}\right|=\sqrt{\mathrm{21,878} {\mathrm{km}}^2}$

$\left|\overrightarrow{d_T}\right|=148 \mathrm{km}$

$\tan \mathrm{\theta }=\frac{94\mathrm{km}}{114.2 \mathrm{km}}$

$\tan \mathrm{\theta }=0.823$

$\mathrm{\theta }={39.4}^{\circ }$

Therefore, the displacement of the streetcar is 148 km [N39°W].