Question

The engines of a plane are pushing it due north at a rate of 300 mph, and the wind is pushing the plane 20° west of north at a rate of 40 mph. In what direction is the plane going?Round to the nearest tenth.

Answer 1

ANSWER:

3° west of north

STEP-BY-STEP EXPLANATION:

To better understand the question we can do the following sketch:

The first thing is to calculate the magnitude of the resulting vector, by means of the law of cosines, as follows:

$\begin{gathered} V_r=\sqrt{(V_1)^2+(V_2)^2-2\cdot V_1\cdot V_2\cdot\cos(\theta)} \\ \\ \text{ We replacing:} \\ \\ V_r=\sqrt{300^2+40^2-2\cdot300\cdot40\cdot\:\cos\left(20°\right)} \\ \\ V_r=262.77 \end{gathered}$

Now, we can calculate the angle, using the law of sines, like this:

$\begin{gathered} \frac{\sin\alpha}{40}=\frac{\sin20\degree}{262.8} \\ \\ \sin\alpha=\frac{\sin20\degree}{262.8}\cdot40 \\ \\ \alpha=\sin^{-1}(0.0520) \\ \\ \alpha=2.98\degree=3.0\degree \end{gathered}$

The direction is 3° west of north