Chapter 14.1, Problem 22WE

To determine

To calculate: The magnitude $\mathrm{w}$ to three significant digits and its bearing to the nearest tenth of degree if $3\left(\mathrm{w}+\mathrm{u}\right)=2\left(\mathrm{w}-\mathrm{v}\right)$ .

Answer to Problem 22WE

The magnitude of $\mathrm{w}\text{ is 2186}$ and its bearing is ${16.8}^{\circ }$ .

Explanation of Solution

Given information:

  $\begin{array}{l}\mathrm{u}:{\text{ magnitude 621, bearing 305}}^{\circ }\\ \mathrm{v}:{\text{ magnitude 336, bearing 15}}^{\circ }\end{array}$

Calculation:

Consider the equation for $\mathrm{w}$ .

  $\begin{array}{c}3\left(\mathrm{w}+\mathrm{u}\right)=2\left(\mathrm{w}-\mathrm{v}\right)\\ 3\mathrm{w}+3\mathrm{u}=2\mathrm{w}-2\mathrm{v}\\ 3\mathrm{w}-2\mathrm{w}=-3\mathrm{u}-2\mathrm{v}\\ \mathrm{w}=-3\mathrm{u}-2\mathrm{v}\end{array}$

Considerthe given information:

  $\begin{array}{l}\mathrm{u}:{\text{ magnitude 621, bearing 305}}^{\circ }\\ \mathrm{v}:{\text{ magnitude 336, bearing 15}}^{\circ }\end{array}$

The angle is:

  $\begin{array}{c}\mathrm{\alpha }=305°-180°-15°\\ \mathrm{\alpha }=110°\end{array}$ .

Now solve for cos law.

  $\begin{array}{c}{\mathrm{w}}^2={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{a}\mathrm{b}\cos \mathrm{\theta }\\ {\mathrm{w}}^2={\left[3\left(621\right)\right]}^2+{\left[2\left(336\right)\right]}^2-2\left(3\left(621\right)\right)\left(2\left(336\right)\right)\cos 110°\\ {\mathrm{w}}^2={\left(1863\right)}^2+{\left(672\right)}^2-2\left(1863\right)\left(672\right)\cos 110°\\ {\mathrm{w}}^2=4778596\\ \mathrm{w}=\sqrt{4778596}\\ \mathrm{w}=2186\end{array}$

Now apply $\sin$ law,

  $\frac{\sin \mathrm{\theta }}{672}=\frac{\sin 110°}{2186}$

Now apply the values of $\sin 110°=0.9396926208$:

  $\frac{\sin \mathrm{\theta }}{672}=\frac{0.9396926208}{2186}$

By cross multiplication:

  $\begin{array}{c}\sin \mathrm{\theta }=0.289\\ \mathrm{\theta }=16.8\end{array}$

The magnitude of $\mathrm{w}\text{ is 2186}$ and its bearing is ${16.8}^{\circ }$ .