## Precalculus (6th Edition) Blitzer

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