Chapter 14.1, Problem 4P

To determine

To calculate: The magnitude to three significant digits and bearings to the nearest tenth of a degree and to find the heading and speed should a plane travel.

A fly due north at $580$ km/h when there is a $45$ km wind blowing from ${130}^{\circ }$ ,

Answer to Problem 4P

The speed and the headingis $581.77km/h,3.48$

Explanation of Solution

Given information:

A fly due north at $580$ km/h when there is a $45$ km wind blowing from $130$ ,

Formula used:

The cosine law:

  $a^2=b^2+c^2-2bc\cos A$

Calculation:

Consider the A fly due north at $580$ km/h when there is a $45$ km wind blowing from $130$ ,

North is a direction of $360\text{ or 0}$ .

Then wind interior from north is $180-130=50$ .

Now by the low of cosines it is written as:

  $a^2=b^2+c^2-2bc\cos A$

So it can be written as:

  $\begin{array}{c}{R}^2={580}^2+{45}^2-2\left(580\right)\left(45\right)cos50\\ R^2=336400+2025-52200\cdot \cos 50\\ R=567.14\text{ km/h}\end{array}$

Now solve for $\sin$ .

  $\frac{\sin \theta }{45}=\frac{\sin 50}{567.14}$

Now apply the values of $\sin 50=0.7660$ :

  $\frac{\sin \theta }{45}=\frac{0.7660}{582}$

By cross multiplication:

  $\begin{array}{c}\frac{\sin \theta }{45}=\frac{0.7660}{582}\\ \sin \theta =0.060782\\ \theta \approx {3.48}^{\circ }\end{array}$

Thus, the speed and heading is $581.77km/h,3.48$.