(a) Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a direction 40.0º north of east (which is equivalent to subtracting B from A —that is, to finding R’=A−B ). (b) Repeat the problem two problems prior, but now you first walk 20.0 m in a direction 40.0º south of west and then 12.0 m in a direction 20.0º east of south (which is equivalent to subtracting A from B —that is, to finding R”=B−A=−R’ ). Show that this is the case.
Solution:
Refer to this problem.
Part A
Consider Figure 3-7A
First, we solve for the value of α using a simple geometry.
\alpha = 40^\circ+\left( 90^\circ-20^\circ \right) = 110^\circ
We can solve for R using the cosine law.
\begin{align*}
R^2 & = A^2 +B^2-2AB \cos \alpha \\
R & = \sqrt{A^2 +B^2-2AB \cos \alpha} \\
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 110^\circ} \\
R & = 26.6115 \ \text{m} \\
R & = 26.6 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Before we can solve for θ, we need to solve for 𝛽 first using the sine law.
\begin{align*}
\frac{\sin \beta}{B} & = \frac{\sin \alpha }{R} \\
\sin \beta & = \frac{B \ \sin \alpha}{R} \\
\beta & = \sin ^{-1} \left( \frac{B \ \sin \alpha}{R} \right) \\
\beta & = \sin ^{-1} \left( \frac{20.0 \ \text{m}\ \sin 110^{\circ}}{26.6115 \ \text{m}} \right) \\
\beta & = 44.9290^\circ
\end{align*}Now, we can solve for θ.
\begin{align*}
\theta & = \left( 90^\circ + 20^\circ \right)-\beta \\
\theta & = 110^\circ - 44.9290^\circ \\
\theta & = 65.071 \\
\theta & = 65.1 ^\circ
\end{align*}Therefore, the compass reading is
65.1^\circ, \text{North of East} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}Part B
Refer to Figure 3-7B
First, we solve for the value of α using a simple geometry.
\alpha = 40^\circ+\left( 90^\circ-20^\circ \right) = 110^\circ
We can solve for R using the cosine law.
\begin{align*}
R^2 & = A^2 +B^2-2AB \cos \alpha \\
R & = \sqrt{A^2 +B^2-2AB \cos \alpha} \\
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 110^\circ} \\
R & = 26.6115 \ \text{m} \\
R & = 26.6 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Before we can solve for θ, we need to solve for 𝛽 first using the sine law.
\begin{align*}
\frac{\sin \beta}{A} & = \frac{\sin \alpha }{R} \\
\sin \beta & = \frac{A \ \sin \alpha}{R} \\
\beta & = \sin ^{-1} \left( \frac{A \ \sin \alpha}{R} \right) \\
\beta & = \sin ^{-1} \left( \frac{12.0 \ \text{m}\ \sin 110^{\circ}}{26.6115 \ \text{m}} \right) \\
\beta & = 25.0708^\circ
\end{align*}Now, we can solve for θ.
\begin{align*}
\theta & = 40^\circ + \beta \\
\theta & = 40^\circ + 25.0708^\circ\\
\theta & = 65.0708^\circ\\
\theta & = 65.1 ^\circ
\end{align*}Therefore, the compass reading is
65.1^\circ, \text{South of West} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}This is consistent with Part A because (A-B) = -(B-A).
Advertisements
Advertisements