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.)
Solution:
Consider Figure 3.5A shown below.
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
\beta = 20^\circ +\left( 90^\circ -40^\circ \right) = 70^\circ
We can use cosine law to solve for R.
\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.
\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.
\theta=74.6488 ^\circ -70 ^\circ = 4.65^\circ
Therefore, the compass reading is
4.65^\circ, \text{South of West} \ \qquad \ {\color{Orange} \left( \text{Answer} \right)}
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