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Solution:
For the four-sided plot to be closed, the resultant displacement of the four sides should be zero. The sum of the horizontal components should be zero, and the sum of the vertical components should also be equal to zero.
We need to solve for the components of each vector. Take into consideration that rightward and upward components are positive, while the reverse is negative.
For vector A, the components are
\begin{align*}
A_x & = \left( 4.70 \ \text{km} \right) \cos 7.5^\circ \\
A_x & = 4.6598 \ \text{km}
\end{align*}\begin{align*}
A_y & = -\left( 4.70 \ \text{km} \right) \sin 7.5^\circ \\
A_y & = -0.6135 \ \text{km}
\end{align*}The components of vector B are
\begin{align*}
B_x & =-\left( 2.48 \ \text{km} \right) \sin 16^\circ \\
B_x & = -0.6836 \ \text{km}
\end{align*}\begin{align*}
B_y & =\left( 2.48 \ \text{km} \right) \cos 16^\circ \\
B_y & =2.3839 \ \text{km}
\end{align*}For vector C, the components are
\begin{align*}
C_x & = -\left( 3.02 \ \text{km} \right) \cos 19^\circ \\
C_x & = -2.8555 \ \text{km}
\end{align*}\begin{align*}
C_y & = \left( 3.02 \ \text{km} \right) \sin 19^\circ \\
C_y & = 0.9832 \ \text{km}
\end{align*}Now, we need to take the sum of the x-components and equate it to zero. The x-component of D is unknown.
\begin{align*}
A_x+B_x+C_x+D_x & =0 \\
4.6598 \ \text{km}-0.6836 \ \text{km}-2.8555 \ \text{km}+ D_x & =0 \\
1.1207 \ \text{km} +D_x & =0 \\
D_x & = -1.1207 \ \text{km}
\end{align*}We also need to take the sum of the y-component and equate it to zero to solve for the y-component of D.
\begin{align*}
A_y +B_y+C_y+D_y & =0 \\
-0.6135 \ \text{km}+2.3839 \ \text{km}+0.9832 \ \text{km}+ D_y & =0 \\
2.7536 \ \text{km} +D_y & =0 \\
D_y & = -2.7536 \ \text{km}
\end{align*}To solve for the distance of D, we shall use the Pythagorean Theorem.
\begin{align*}
D & = \sqrt{\left( D_x \right)^2+\left( D_y \right)^2} \\
D & = \sqrt{\left( -1.1207 \ \text{km} \right)^2+\left( -2.7536 \ \text{km} \right)^2} \\
D & = 2.97 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Then we can solve for θ using the tangent function. Since it is taken from the vertical axis, it can be solved by:
\begin{align*}
\theta & = \tan^{-1} \left| \frac{D_x}{D_y} \right|
\\
\theta & = \tan^{-1} \left| \frac{-1.1207 \ \text{km}}{-2.7536 \ \text{km}} \right|
\\
\theta & = 22.1 ^ \circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Solution:
Part A
Consider the illustration shown.
The south and west components of the 32.0 km distance are denoted by DS and DW, respectively. The values of these components are solved below:
\begin{align*}
D_S & = \left( 32.0\ \text{km} \right) \sin 35.0 ^\circ \\
D_S & = 18.4^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}\begin{align*}
D_W & = \left( 32.0\ \text{km} \right) \cos 35.0 ^\circ \\
D_W & = 26.2^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Part B
Consider the new set of axes (X-Y) as shown below. This new set of axes is rotated 45° from the original axes. Thus, axis X is 45° south of west, and axis Y is 45° west of north. First, we can obviously see that θ has a value of 10°.
Therefore, the components of the 32.0 km distance along X and Y axes are:
\begin{align*}
D_X & = \left( 32.0 \ \text{km} \right) \cos 10^\circ \\
D_X & = 31.5^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}\begin{align*}
D_Y & = \left( 32.0 \ \text{km} \right) \sin 10^\circ \\
D_Y & = 5.56^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Solution:
Consider the illustration shown.
We need to solve for an interior angle of the triangle. So, we need to solve for the value of α first. This can be done by simply subtracting the sum of 21 and 11 degrees from 90 degrees.
\begin{align*}
\alpha & = 90 ^ \circ -\left( 21^\circ +11^\circ \right) \\
\alpha & = 58^\circ
\end{align*}Then, using the cosine law, we can now solve for the magnitude of vector C. That is
\begin{align*}
C^2 & = A^2 + B^2 - 2AB \cos \alpha \\
C^2 & = \left( 80\ \text{m} \right)^2+\left( 105\ \text{m} \right)^2-2\left( 80\ \text{m} \right)\left( 105\ \text{m} \right) \cos 58^\circ \\
C^2 & = 8522.3564 \\
C & = \sqrt{8522.3564} \\
C & = 92.3 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Before we can solve for the value of θ, we need to know the value of β first. This can be done by using the sine law.
\begin{align*}
\frac{\sin \beta}{80\ \text{m}} & = \frac{\sin 58^\circ }{92.3 \ \text{m}} \\
\sin \beta & = \frac{\left( 80 \ \text{m} \right)\sin 58^\circ }{92.3 \ \text{m}} \\
\beta & = \arcsin \left[ \frac{\left( 80 \ \text{m} \right)\sin 58^\circ }{92.3 \ \text{m}} \right] \\
\beta & = 47.3^\circ
\end{align*}Finally, we can solve for θ.
\begin{align*}
\theta & = \left( 90 ^\circ +11^\circ \right) - \beta \\
\theta & = \left( 90 ^\circ +11^\circ \right) - 47.3^\circ \\
\theta & = 53.7^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Solution:
Part A
Consider the illustration shown.
Let DE be the east component of the distance, and DN be the north component of the distance.
\begin{align*}
D_E & = 7.50 \ \sin 15^\circ \\
D_E & = 1.9411 \ \text{km} \\
D_E & = 1.94 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}\begin{align*}
D_N & = 7.50 \ \cos 15^\circ \\
D_N & = 7.2444\ \text{km} \\
D_N & = 7.24 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Part B
It can be obviously seen from the figure below that you still arrive at the same point if the east and north legs are reversed in order.
Solution:
Considering the right triangle formed by the vectors A, B, and R. We can solve for the magnitude of R using the Pythagorean Theorem. That is
\begin{align*}
R & = \sqrt{A^2+B^2} \\
& = \sqrt{\left( 18.0 \text{m} \right)^2+\left( 25.0 \text{m} \right)^2} \\
& =30.806 \text{m} \\
& \approx 30.8 \text{m} \qquad {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Then we solve for the compass direction by solving the value θ using the same right triangle.
\begin{align*}
\theta & = \arctan \left( \frac{B}{A} \right) \\
& = \arctan \left( \frac{25.0 \text{m}}{18.0 \text{m}} \right) \\
& = 54.246 ^\circ \\
& \approx 54.2 ^ \circ \\
\end{align*}Therefore, the compass direction of the resultant is 54.2° North of West.
Solution:
Considering the right triangle formed by the vectors A, B, and R. We can solve for the magnitude of R using the Pythagorean Theorem. That is
\begin{align*}
R & = \sqrt{A^2+B^2} \\
& = \sqrt{\left( 18.0\ \text{m} \right)^2+\left( 25.0\ \text{m} \right)^2} \\
& =30.806 \ \text{m} \\
& \approx 30.8 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Then we solve for the compass direction by solving the value θ using the same right triangle.
\begin{align*}
\theta & = \arctan\left( \frac{B}{A} \right) \\
& = \arctan\left( \frac{25.0\ \text{m}}{18.0\ \text{m}} \right) \\
& = 54.246 ^\circ \\
& \approx 54.2 ^ \circ
\end{align*}Therefore, the compass direction of the resultant is 54.2° North of West.
Solution:
Consider the following figure.
Using the right triangle formed, we can solve for the east component and the north component. Let SE be the east component and SN be the north component of S.
\begin{align*}
S_E & = S \cos 45^\circ \\
& = \left( 123 \ \text{km} \right) \cos45^\circ \\
& = 86.974 \ \text{km} \\
& \approx 87.0 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}\begin{align*}
S_N & = S \sin 45^\circ \\
& = \left( 123 \ \text{km} \right) \sin 45^\circ \\
& = 86.974 \ \text{km} \\
& \approx 87.0 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Therefore, the east and north components are equal at about 87.0 km.
Solution:
Part A
Looking at path D, we can see that it moves 2 blocks downward, 6 blocks to the right, 4 blocks upward, and 1 block to the left. Thus, the total distance of path D is
\begin{align*}
\text{distance} & = \left( 2\times 120\ \text{m} \right)+\left( 6\times 120\ \text{m} \right)+\left( 4\times 120\ \text{m} \right)+\left( 1\times 120\ \text{m} \right) \\
& = 1\ 560 \ \text{m} \\
& = 1.56 \times 10^{3} \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Part B
Looking at the initial and final position of path D, the final position is 5 blocks to the right or 600 meters to the right of the initial position, and 2 blocks or 240 meters upward from the initial position. Refer to the figure below.
Using the right triangle, we can solve for the displacement using the Pythagorean Theorem.
\begin{align*}
\text{displacement} & = \sqrt{\left( 600\ \text{m} \right)^2+\left( 240\ \text{m} \right)^2} \\
& = 646.2198 \ \text{m} \\
& \approx 646 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Solution:
Part A
Looking at path C, it moves 1 block upward, 5 blocks to the right, 2 blocks downward, 1 block to the left, 1 block upward, and 3 blocks to the left. So, the total distance is
\begin{align*}
\text{distance} \ = \ &\left( 1\times 120 \ \text{m} \right)+\left( 5\times 120\ \text{m} \right)+\left( 2\times 120\ \text{m} \right)+\left( 1\times 120\ \text{m} \right) \\
& +\left( 1\times 120\ \text{m} \right)+\left( 3\times 120\ \text{m} \right) \\
= \ & 120\ \text{m}+600 \ \text{m}+240 \ \text{m} + 120 \ \text{m}+ 120 \ \text{m}+360\ \text{m} \\
= \ & 1560 \ \text{m} \ \qquad \ {\color{DarkOrange}\left( \text{Answer} \right) }
\end{align*}Part B
It can be seen from the figure that the end of path C is just one block to the right from the starting point. Therefore, the magnitude of the displacement is
\begin{align*}
\text{displacement} = 120\ \text{m} \ \qquad \ {\color{DarkOrange}\left( \text{Answer} \right) }
\end{align*}The direction is to the right or is equivalent to 0° measured from the positive x-axis.
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