Chapter 14.1, Problem 3P

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 due east,

Answer to Problem 3P

The speed and headingis ${0.4411}^{\circ },581.77km/h$

Explanation of Solution

Given information:

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

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 due east,

Now by the low of cosines it is written as:

  $H^2=B^2+P^2-2BP\cos 90$

So it can be written as:

  $\begin{array}{c}{R}^2={580}^2+{45}^2-2\left(580\right)\left(45\right)cos90\\ R^2=336400+2025-52200\times 0\\ R=581.77\text{ km/h}\end{array}$

Now solve for $\sin$ .

  $\frac{\sin \theta }{45}=\frac{\sin 90}{582}$

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

  $\frac{\sin \theta }{45}=\frac{1}{582}$

By cross multiplication:

  $\begin{array}{c}\frac{\sin \theta }{45}=\frac{1}{582}\\ \sin \theta =\frac{45}{582}\\ \theta ={\sin }^{-1}\left(\frac{45}{582}\right)\\ \theta \approx {0.4411}^{\circ }\end{array}$

Thus, thespeed and heading is ${0.4411}^{\circ },581.77km/h$