Chapter 14.1, Problem 19WE

To determine

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

  $\begin{array}{l}u:{\text{ magnitude 460, bearing 0}}^{\circ }\\ v:{\text{ magnitude 712, bearing 130}}^{\circ }\\ w:u+v\end{array}$

Answer to Problem 19WE

Themagnitude of $w\text{ is 545}$ and its bearing is ${90}^{\circ }$ .

Explanation of Solution

Given information:

  $\begin{array}{l}u:{\text{ magnitude 460, bearing 0}}^{\circ }\\ v:{\text{ magnitude 712, bearing 130}}^{\circ }\\ w:u+v\end{array}$

Calculation:

Considerthe given information:

  $\begin{array}{l}u:{\text{ magnitude 460, bearing 0}}^{\circ }\\ v:{\text{ magnitude 712, bearing 130}}^{\circ }\\ w:u+v\end{array}$

The angle is:

  $\begin{array}{c}\alpha =180-130\\ \alpha =50\end{array}$ .

Now solve for cos law.

  $\begin{array}{c}{w}^2={460}^2+{712}^2-2\left(460\right)\left(712\right)\cos 50\\ w^2=211600+506944-655040\times 0.64278760\\ w^2=718544-421051.5895\\ w=\sqrt{297492.4141}\\ w=545\end{array}$

Now,

  $\frac{\sin \theta }{712}=\frac{\sin 80}{545}$

Now apply the values of $\sin 50=0.64278760$ :

  $\frac{\sin \theta }{712}=\frac{0.64278760}{545}$

By cross multiplication:

  $\begin{array}{c}\sin \theta =1.0\\ \theta =90\end{array}$

The magnitude of $w\text{ is 545}$ and its bearing is ${90}^{\circ }$ .

Thus, the magnitude of $w\text{ is 545}$ and its bearing is ${90}^{\circ }$ .