Chapter 14.1, Problem 20WE

To determine

To calculate: The magnitude $w$ to three significant digits and its bearing to the nearest tenth of degree.

  $\begin{array}{l}u:\text{ magnitude 23}{\text{.0, bearing 215}}^{\circ }\\ v:\text{ magnitude 14}{\text{.5, bearing 105}}^{\circ }\\ w:u+v\end{array}$

Answer to Problem 20WE

Themagnitude of $w\text{ is 22}\text{.6}$ and its bearing is ${37.1}^{\circ }$ .

Explanation of Solution

Given information:

  $\begin{array}{l}u:\text{ magnitude 23}{\text{.0, bearing 215}}^{\circ }\\ v:\text{ magnitude 14}{\text{.5, bearing 105}}^{\circ }\\ w:u+v\end{array}$

Calculation:

Considerthe given information:

  $\begin{array}{l}u:\text{ magnitude 23}{\text{.0, bearing 215}}^{\circ }\\ v:\text{ magnitude 14}{\text{.5, bearing 105}}^{\circ }\\ w:u+v\end{array}$

The angle is:

  $\begin{array}{c}\alpha =105-35\\ \alpha =70\end{array}$ .

Now solve for cos law.

  $\begin{array}{c}{w}^2={23.0}^2+{14.5}^2-2\left(23.0\right)\left(14.5\right)\cos 70\\ w^2=529+210.25-667\times 0.34202014463751\\ w^2=739.25-228.1274365\\ w=\sqrt{511.1225635}\\ w=22.6\end{array}$

Now apply $\sin$ law,

  $\frac{\sin \theta }{14.5}=\frac{\sin 70}{22.6}$

Now apply the values of $\sin 70=0.34202014463751$ :

  $\frac{\sin \theta }{14.5}=\frac{0.34202014463751}{22.6}$

By cross multiplication:

  $\begin{array}{c}\sin \theta =0.603\\ \theta =37.1\end{array}$

The magnitude of $w\text{ is 22}\text{.6}$ and its bearing is ${37.1}^{\circ }$ .

Thus, the magnitude of $w\text{ is 22}\text{.6}$ and its bearing is ${37.1}^{\circ }$ .