Repeat the problem above, but reverse the order of the two legs of the walk; show that you get the same final result. That is, you first walk leg B , which is 20.0 m in a direction exactly 40º south of west, and then leg A , which is 12.0 m in a direction exactly 20º west of north. (This problem shows that A+B=B+A.)
Solution:
Consider Figure 3-6A below with B drawn first before A.
Compute for the value of angle β by adding 20° and the complement of 40°. This is by simple geometry.
\begin{align*}
\beta & = 20^\circ +\left( 90^\circ -40^\circ \right) \\
\beta & = 70^\circ \\
\end{align*}Solve for the magnitude of R using cosine law.
\begin{align*}
R^2 & = A^2 +B^2 - 2 A B \cos \beta \\
R & = \sqrt{A^2 +B^2 - 2 A B \cos \beta} \\
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*}To solve for θ, we need to solve angle α first. This can be done using the sine law.
\begin{align*}
\frac{\sin \alpha}{A} & = \frac{\sin \beta }{R} \\
\frac{\sin \alpha}{12.0 \ \text{m}} & = \frac{\sin 70^\circ }{19.4892 \ \text{m}} \\
\sin \alpha & = \frac{12.0 \ \text{m}\ \sin 70^\circ}{19.4892 \ \text{m}} \\
\alpha & = \sin ^{-1} \left( \frac{12.0 \ \text{m}\ \sin 70^\circ}{19.4892 \ \text{m}} \right) \\
\alpha & = 35.3516 ^ \circ
\end{align*}Finally, we can solve for θ.
\begin{align*}
\theta & = 40^\circ - 35.3516^\circ \\
& = 4.6484 ^\circ \\
& = 4.65 ^\circ
\end{align*}Therefore, the compass reading is
4.65^\circ, \text{South of West} \ \qquad \ {\color{Orange} \left( \text{Answer} \right)}This is the same with Problem 5.
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