(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
Figure 3-7A First, we solve for the value of α
α = 4 0 ∘ + ( 9 0 ∘ − 2 0 ∘ ) = 11 0 ∘ \alpha = 40^\circ+\left( 90^\circ-20^\circ \right) = 110^\circ α = 4 0 ∘ + ( 9 0 ∘ − 2 0 ∘ ) = 11 0 ∘ We can solve for R
R 2 = A 2 + B 2 − 2 A B cos α R = A 2 + B 2 − 2 A B cos α R = ( 12.0 m ) 2 + ( 20.0 m ) 2 − 2 ( 12.0 m ) ( 20.0 m ) cos 11 0 ∘ R = 26.6115 m R = 26.6 m ( Answer ) \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*} R 2 R R R R = A 2 + B 2 − 2 A B cos α = A 2 + B 2 − 2 A B cos α = ( 12.0 m ) 2 + ( 20.0 m ) 2 − 2 ( 12.0 m ) ( 20.0 m ) cos 11 0 ∘ = 26.6115 m = 26.6 m ( Answer ) Before we can solve for θ 𝛽 first using the sine law.
sin β B = sin α R sin β = B sin α R β = sin − 1 ( B sin α R ) β = sin − 1 ( 20.0 m sin 11 0 ∘ 26.6115 m ) β = 44.929 0 ∘ \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*} B sin β sin β β β β = R sin α = R B sin α = sin − 1 ( R B sin α ) = sin − 1 ( 26.6115 m 20.0 m sin 11 0 ∘ ) = 44.929 0 ∘ Now, we can solve for θ
θ = ( 9 0 ∘ + 2 0 ∘ ) − β θ = 11 0 ∘ − 44.929 0 ∘ θ = 65.071 θ = 65. 1 ∘ \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*} θ θ θ θ = ( 9 0 ∘ + 2 0 ∘ ) − β = 11 0 ∘ − 44.929 0 ∘ = 65.071 = 65. 1 ∘ Therefore, the compass reading is
65. 1 ∘ , North of East ( Answer ) 65.1^\circ, \text{North of East} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} 65. 1 ∘ , North of East ( Answer ) Part B
Refer to Figure 3-7B
Figure 3-7B First, we solve for the value of α
α = 4 0 ∘ + ( 9 0 ∘ − 2 0 ∘ ) = 11 0 ∘ \alpha = 40^\circ+\left( 90^\circ-20^\circ \right) = 110^\circ α = 4 0 ∘ + ( 9 0 ∘ − 2 0 ∘ ) = 11 0 ∘ We can solve for R
R 2 = A 2 + B 2 − 2 A B cos α R = A 2 + B 2 − 2 A B cos α R = ( 12.0 m ) 2 + ( 20.0 m ) 2 − 2 ( 12.0 m ) ( 20.0 m ) cos 11 0 ∘ R = 26.6115 m R = 26.6 m ( Answer ) \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*} R 2 R R R R = A 2 + B 2 − 2 A B cos α = A 2 + B 2 − 2 A B cos α = ( 12.0 m ) 2 + ( 20.0 m ) 2 − 2 ( 12.0 m ) ( 20.0 m ) cos 11 0 ∘ = 26.6115 m = 26.6 m ( Answer ) Before we can solve for θ 𝛽 first using the sine law.
sin β A = sin α R sin β = A sin α R β = sin − 1 ( A sin α R ) β = sin − 1 ( 12.0 m sin 11 0 ∘ 26.6115 m ) β = 25.070 8 ∘ \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*} A sin β sin β β β β = R sin α = R A sin α = sin − 1 ( R A sin α ) = sin − 1 ( 26.6115 m 12.0 m sin 11 0 ∘ ) = 25.070 8 ∘ Now, we can solve for θ
θ = 4 0 ∘ + β θ = 4 0 ∘ + 25.070 8 ∘ θ = 65.070 8 ∘ θ = 65. 1 ∘ \begin{align*}
\theta & = 40^\circ + \beta \\
\theta & = 40^\circ + 25.0708^\circ\\
\theta & = 65.0708^\circ\\
\theta & = 65.1 ^\circ
\end{align*} θ θ θ θ = 4 0 ∘ + β = 4 0 ∘ + 25.070 8 ∘ = 65.070 8 ∘ = 65. 1 ∘ Therefore, the compass reading is
65. 1 ∘ , South of West ( Answer ) 65.1^\circ, \text{South of West} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} 65. 1 ∘ , South of West ( Answer ) This is consistent with Part A because (A-B) = -(B-A) .
