Chapter 14.1, Problem 18WE

To determine

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

  $\begin{array}{l}u:{\text{ magnitude 1850, bearing 125}}^{\circ }\\ v:{\text{ magnitude 2960, bearing 25}}^{\circ }\\ w:u+2v\end{array}$

Answer to Problem 18WE

Themagnitude of $w\text{ is 3207}$ and its bearing is ${65.4}^{\circ }$ .

Explanation of Solution

Given information:

  $\begin{array}{l}u:{\text{ magnitude 1850, bearing 125}}^{\circ }\\ v:{\text{ magnitude 2960, bearing 25}}^{\circ }\\ w:u+2v\end{array}$

Calculation:

Considerthe given information:

  $\begin{array}{l}u:{\text{ magnitude 1850, bearing 125}}^{\circ }\\ v:{\text{ magnitude 2960, bearing 25}}^{\circ }\\ w:u+2v\end{array}$

The angle is:

  $\begin{array}{c}\alpha =55+25\\ \alpha =80\end{array}$ .

Now solve for cos law.

  $\begin{array}{c}{w}^2={1850}^2+{2960}^2-2\left(1850\right)\left(2960\right)\cos 80\\ w^2=3422500+8761600-10952000\times 0.1736481\\ w^2=12184100-1901793.991\\ w=\sqrt{10282306.01}\\ w=3207\end{array}$

Now,

  $\frac{\sin \theta }{2960}=\frac{\sin 80}{3207}$

Now apply the values of $\sin 80=0.1736481$ :

  $\frac{\sin \theta }{2960}=\frac{0.1736481}{3207}$

By cross multiplication:

  $\begin{array}{c}\sin \theta =0.909\\ \theta =65.4\end{array}$

The magnitude of $w\text{ is 3207}$ and its bearing is ${65.4}^{\circ }$ .

Thus, the magnitude of $w\text{ is 3207}$ and its bearing is ${65.4}^{\circ }$ .