Question

A jet leaves a runway whose bearing is N 36°E from the control tower. After flying 4 miles, the jet turns 90° and flies on a bearing of S 54°E for 5 miles. At that time, what is the bearing of the jet from the control tower?

Answer 1

The initial bearing is N 36° E. Then, the jet turns 90° after 4 miles and flies on a bearing of S 54° E for 5 miles. The diagram of the problem is:

To solve this, we need to know the angle x. First of all, we calculate the distance traveled vertically and horizontally in the first part:

`\begin{gathered} x_1=4\cdot\sin 36\degree \\ y_1=4\cdot\cos 36\degree \end{gathered}`

Now, for the second part:

`\begin{gathered} x_2=5\sin 54\degree \\ y_2=5\cos 54\degree \end{gathered}`

As we can see from the figure, the jet first travels to the NE and then to the SE, so the horizontal net distance is the sum of the two parts, and the vertical net distance is the subtraction:

`\begin{gathered} x=x_1+x_2=4\sin 36\degree+5\sin 54\degree\approx6.3962^{} \\ y=y_1-y_2=4\cos 36\degree-^{}5\cos 54\degree\approx0.2971 \end{gathered}`

The angle of the final position and the horizontal line is given by:

`\theta=\tan ^(-1)((0.2971)/(6.3962))=2.6594\degree\approx3\degree`

The complement of this angle is the angle we are looking for, that is:

`90\degree-3\degree=87\degree`

Since the first angle is positive, then our direction will be NE, then the bearing of the jet from the control tower will be:

`N87\degree E`