Suppose you walk 18.0 m straight west and then 25.0 m straight north. 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 \vec{A} and \vec{B} , as in Figure 3.53, then this problem asks you to find their sum \vec{R}=\vec{A}+\vec{B} .)
Solution:
Consider Figure 3.54A.
The resultant of the two vectors \vec{A} and \vec{B} is labeled \vec{R}. This \vec{R} is directed \theta ^{\circ} from the x-axis.
We shall use the right triangle formed to solve for the unknowns.
Solve for the magnitude of the resultant.
\begin{align*} R & = \sqrt{A^2 +B^2} \\ R & = \sqrt{\left(18.0 \ \text{m} \right)^2+\left( 25.0 \ \text{m} \right)^2} \\ R & = 30.8 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
Solve for the value of \theta .
\begin{align*} \theta & = \arctan \left( \frac{B}{A} \right) \\ \theta & = \arctan \left( \frac{25.0 \ \text{m}}{18.0 \ \text{m}} \right) \\ \theta & = 54.2^\circ \end{align*}
We need the complementary angle for the compass angle.
\begin{align*} 90^\circ -54.2^\circ =35.8^\circ \end{align*}
Therefore, the compass angle reading is
\begin{align*} 35.8^\circ , \text{W of N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
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