Example 17 from Algebra and Trigonometry, 10.8 Vectors

Finding Ground Speed and Bearing Using Vectors

We now have the tools to solve the problem we introduced in the opening of the section.

An airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140°. A north wind (from north to south) is blowing at 16.2 miles per hour. What are the ground speed and actual bearing of the plane? See Figure 19.

Image of a plan flying SE at 140 degrees and the north wind blowing.

Figure 19

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
The ground speed is represented by $x$ in the diagram, and we need to find the angle $α$ in order to calculate the adjusted bearing, which will be $140° + α .$
Notice in Figure 19, that angle $b C O$ must be equal to angle $A O C$ by the rule of alternating interior angles, so angle $b C O$ is 140°. We can find $x$ by the Law of Cosines:
$\begin{array}{l} x^{2} = {(16.2)}^{2} + {(200)}^{2} - 2 (16.2) (200) \cos (140°) \\ x^{2} = 45, 226.41 \\ x = \sqrt{45, 226.41} \\ x = 212.7 \end{array}$
The ground speed is approximately 213 miles per hour. Now we can calculate the bearing using the Law of Sines.

$\begin{array}{l} \frac{\sin α}{16.2} = \frac{\sin (140°)}{212.7} \\ \sin α = \frac{16.2 \sin (140°)}{212.7} \\ = 0.04896 \\ \sin^{- 1} (0.04896) = 2.8° \end{array}$
Therefore, the plane has a SE bearing of 140°+2.8°=142.8°. The ground speed is 212.7 miles per hour.

How did it go?

Solution

Solution