A helicopter flies from the airport on a course with a bearing of 15 degrees.

Question

After flying for 108 ​miles, the helicopter flies due east for some time. The helicopter flies back to the airport with a bearing of 228 degrees. How far did the helicopter fly on the final leg of its​ journey?

Answer 1

A question with bearings and trigonometry, we should start by reminding ourselves that bearings are clockwise from North (0 deg)... and by making a little sketch so we know what our problem looks like.

Start at point A. Draw a line on bearing 15 deg to represent 108 miles. The end point is B. This is first leg. AB = 108.

Point C is due East (bearing of 90 deg) from point B. Draw a line BC for this second leg. We don't know the length of BC.

Join C back to A for the third and final leg CA. We also don't know the length of CA.

Your sketch won't be an accurate representation because we don't know some lengths, so don't measure anything, or guess an angle based on your sketch!

However, we do know that we have a triangle with one known side. And we need to find one of the other sides CA the last leg. For triangle problems, we must know 3 things to be able to solve it, so this tells us we should be able to work out 2 of the angles (which means we can work out all 3). So let's do that...

The easiest is the angle BCA, I think. As BC is horizontal (due East on 90 deg bearing), and the homeward bearing for CA is 228 deg. So you can see from your sketch that 228 + BCA + 90 = 360 (degrees in a complete rotation). Hence angle BCA is 42 degrees.

Next, let's look at angle ABC. Because the second leg is due east (90 deg) we can make a right angle triangle AZB where Z is on the North line through A and the East line through B. You might want to add this to your sketch. We know angle ZAB is 15 deg, because this is the initial bearing. So angle ZBA = 75 deg (because angles in a triangle = 180 deg and 180 - 90 - 15 = 75). Because angles on a straight line also equal 180 deg, we can immediately get angle ABC = 105 deg (because it's 180-75). Because angles in a triangle equal 180, then our last angle BAC = 33 deg (180-105-42).

To find the unknown side we can use either the sine rule or the cosine rule. Because we have a side + its angle, and an unknown side + its angle, the sine rule looks like a better fit here.

The sine rule says sin A / a = sin B / b = sin C / c

For our triangle, we can say sin (angle BCA) / BA = sin (angle CBA) / CA

=> sin (42 deg) / 108 = sin (105 deg) / CA

rearrange to get CA the required side on its own

=> CA sin (42 deg) = 108 sin (105 deg)

=> CA = 108 sin (105 deg) / sin (42 deg)

=> CA = 155.9 miles (156 miles to the nearest mile)

If you only know the maths for right angled triangles, you can still work it out, but it takes two steps, as follows:

As we already created a right angle triangle AZB, we can use standard sin/cos/tan to work out the side AZ.

AZ/hypotenuse AB = cos (angle ZAB)

=> AZ / 108 = cos (15 deg)

=> AZ = 108 cos (15 deg)

And now we can use the large right angle triangle AZC to work out the hypotenuse CA which we need.

AZ/hypotenuse CA = cos (angle ZAC)

=> AZ / CA = cos (48 deg)

=> CA = AZ / cos (48 deg)

which as we previously found AZ

=> CA = 108 cos (15 deg) / cos (48 deg)

=> CA = 155.9 miles (or 156 miles to nearest mile)

CA must be more than 108 so it seems plausible. You can use a "triangle calculator" online to double check if you want. As I got the same answer using two different methods, I am quietly confident. :-)