## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Section 6.6 - Vectors - Exercise Set - Page 784: 115

#### Answer

a) $104{}^\circ$ b) Decreases

#### Work Step by Step

a) Let the velocity of the plane with respect to ground be ${{\mathbf{v}}_{1}}$. The velocity of the plane with respect to wind is ${{\mathbf{v}}_{2}}$. The velocity of wind with respect to earth is $\mathbf{u}$. The angle of the plane with respect to wind is \begin{align} & \tan \theta =\frac{{{\mathbf{v}}_{1}}}{\mathbf{u}} \\ & \tan \theta =\frac{310}{75} \\ & \theta =76{}^\circ \end{align} Here, $\theta$ is the angle of the plane’s velocity with respect to the wind. So, the direction of the plane from the east to north at an angle $\left( \gamma \right)$ is \begin{align} & \gamma =180{}^\circ -76{}^\circ \\ & =104{}^\circ \end{align} Hence, the direction for where you should head the plane is $104{}^\circ$ from east to north. (b) The angle of the plane with respect to wind is $\tan \theta =\frac{{{\mathbf{v}}_{1}}}{\mathbf{u}}$ When the airspeed increases, then the value of $\tan \theta$ decreases; the value of $\theta$ also decreases. Maintaining the angle also increases the plane’s speed with respect to the earth. Hence, the direction angle decreases.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.