Chapter 14.1, Problem 15WE

To determine

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

  $\begin{array}{l}u:{\text{ magnitude 117, bearing 130}}^{\circ }\\ v:{\text{ magnitude 102, bearing 220}}^{\circ }\\ w:u+v\end{array}$

Answer to Problem 15WE

Themagnitude of $w\text{ is 155 }$ and its bearing is ${41.2}^{\circ }$ .

Explanation of Solution

Given information:

  $\begin{array}{l}u:{\text{ magnitude 117, bearing 130}}^{\circ }\\ v:{\text{ magnitude 102, bearing 220}}^{\circ }\\ w:u+v\end{array}$

Calculation:

Considerthe given information:

  $\begin{array}{l}u:{\text{ magnitude 117, bearing 130}}^{\circ }\\ v:{\text{ magnitude 102, bearing 220}}^{\circ }\\ w:u+v\end{array}$

The angle is:

  $\begin{array}{c}50+\alpha +40=180\\ \alpha +90=180\\ \alpha =180-90\\ \alpha =90\end{array}$ .

Now solve for cos law.

  $a^2=b^2+c^2-2bc\cos \theta$

As apply the values in the $\cos$ law.

  $\begin{array}{c}{w}^2={117}^2+{102}^2-2\left(117\right)\left(102\right)\cos 90\\ w^2=13689+10404\\ w^2=24093\\ w=\sqrt{24093}\\ w=155\end{array}$

Now,

  $\frac{\sin \theta }{102}=\frac{\sin 90}{155}$

Now apply the values of $\sin 90=1$ :

  $\frac{\sin \theta }{102}=\frac{1}{155}$

By cross multiplication:

  $\begin{array}{c}\sin \theta =0.6580\\ \theta =41.2\end{array}$

The magnitude of $w\text{ is 155 }$ and its bearing is ${41.2}^{\circ }$ .

Thus, the magnitude of $w\text{ is 155 }$ and its bearing is ${41.2}^{\circ }$ .