College Physics by Openstax Chapter 3 Problem 5

Suppose you first walk 12.0 m in a direction 20º west of north and then 20.0 m in a direction 40.0º south of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B, as in Figure 3.54, then this problem finds their sum R=A+B.)

Figure 3.54

Solution:

Consider Figure 3.5A shown below.

Figure 3.5A

Before we can use cosine law to solve for the magnitude of R, we need to solve for the interior angle 𝛽 first. The value of 𝛽 can be calculated by inspecting the figure and use simple knowledge on geometry. It is equal to the sum of 20° and the complement of 40°. That is

β=20∘+(90∘−40∘)=70∘\beta = 20^\circ +\left( 90^\circ -40^\circ \right) = 70^\circ

We can use cosine law to solve for R.

R2=A2+B2−2ABcos⁡βR2=(12.0 m)2+(20.0 m)2−2(12.0 m)(20.0 m)cos⁡70∘R=(12.0 m)2+(20.0 m)2−2(12.0 m)(20.0 m)cos⁡70∘R=19.4892 mR=19.5 m  (Answer)\begin{align*} R^2 & =A^2+B^2 -2AB \cos \beta \\ R^2 & = \left( 12.0\ \text{m} \right) ^2+\left( 20.0\ \text{m} \right)^2-2 \left( 12.0\ \text{m} \right) \left( 20.0\ \text{m} \right) \cos 70^\circ \\ R & = \sqrt{ \left( 12.0\ \text{m} \right) ^2+\left( 20.0\ \text{m} \right)^2-2 \left( 12.0\ \text{m} \right) \left( 20.0\ \text{m} \right) \cos 70^\circ} \\ R & =19.4892 \ \text{m} \\ R & =19.5 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

We can solve for α using sine law.

sin⁡αB=sin⁡βRsin⁡α20.0 m=sin⁡70∘19.4892 msin⁡α=20.0 sin⁡70∘19.4892α=sin⁡−1(20.0 sin⁡70∘19.4892)α=74.6488∘\begin{align*} \frac{\sin \alpha}{B} & = \frac{\sin \beta}{R} \\ \frac{\sin \alpha}{20.0\ \text{m}} & = \frac{\sin 70^\circ }{19.4892 \ \text{m}} \\ \sin \alpha & = \frac{20.0 \ \sin 70^\circ }{19.4892} \\ \alpha & = \sin ^{-1} \left( \frac{20.0 \ \sin 70^\circ }{19.4892} \right) \\ \alpha & = 74.6488 ^\circ \end{align*}

Then we solve for the value of θ by subtracting 70° from α.

θ=74.6488∘−70∘=4.65∘\theta=74.6488 ^\circ -70 ^\circ = 4.65^\circ

Therefore, the compass reading is

4.65∘,South of West  (Answer)4.65^\circ, \text{South of West} \ \qquad \ {\color{Orange} \left( \text{Answer} \right)}
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