Coaching Centre
Compass and true bearings
Writing directions as compass bearings and true bearings, converting between them, back bearings, and finding the angles inside a triangle formed by a two-leg trip.
Pilots, ferry masters, bushwalkers and surveyors all describe direction with an angle measured from north. "Head north-east" is too vague when you are $20$ km out to sea. A bearing pins the direction down to the degree, and once a trip is drawn with bearings, the triangle it makes can be solved with trigonometry.
Two ways to write a bearing
| Compass bearing | True bearing | |
|---|---|---|
| Starts from | north or south | north only |
| Turns | towards east or west, at most $90^\circ$ | clockwise, $0^\circ$ to $360^\circ$ |
| Written as | N $35^\circ$ E, S $40^\circ$ W | three digits and T: $035^\circ$T, $220^\circ$T |
True bearings always use three digits, so $35^\circ$ east of north is written $035^\circ$T. The leading zero stops anyone misreading it, which matters when lives depend on it.
Write each compass bearing as a true bearing: (a) N $35^\circ$ E (b) S $40^\circ$ W (c) N $50^\circ$ W.
Back bearings
If the bearing of $B$ from $A$ is $065^\circ$T, what is the bearing of $A$ from $B$? You are looking back along the same line, so you face the opposite way: add $180^\circ$. If that takes you past $360^\circ$, subtract $180^\circ$ instead.
The bearing of Shark Island from a ferry is $290^\circ$T. What is the bearing of the ferry from Shark Island?
Read the words "from" carefully. "The bearing of $B$ from $A$" means stand at $A$, face north, and turn until you face $B$. The point after "from" is where you stand.
Two-leg trips
When a ship sails on one bearing and then turns onto another, the two legs and the straight line home make a triangle. To use trigonometry you need the angles inside that triangle. The trick is to draw a fresh north line at the turning point, because bearings are always measured from the north line where you are.
Set leg 1 to $050^\circ$T for $5$ km and leg 2 to $140^\circ$T for $3$ km. Check the angle at A and the trip home against Example 4. Then set leg 2 to $090^\circ$T. The angle at A is no longer $90^\circ$: predict it before you look, using the back bearing.
Read the compass form of each leg in the readouts. Find a leg 1 bearing whose compass form starts with S and ends with W.
A ferry sails $8$ km on a bearing of $090^\circ$T, then $6$ km on $180^\circ$T. Find its distance from the start and its bearing from the start, to the nearest degree.
Mei walks $5$ km on a bearing of $050^\circ$T, then turns and walks $3$ km on $140^\circ$T. Find the angle at the turning point, her distance from the start, and her bearing from the start.
When the angle at the turn is not $90^\circ$, Pythagoras no longer works. That is exactly what the cosine rule in 4E is for, and 4F puts bearings and the cosine rule together.
Your turn
Write $235^\circ$T as a compass bearing, and find its back bearing.
Show the working
$235^\circ$ is $55^\circ$ past south ($180^\circ$), turning towards west: S $55^\circ$ W. Back bearing $= 235^\circ - 180^\circ = 055^\circ$T.
The compass bearing S $40^\circ$ W, written as a true bearing, is:
South is $180^\circ$T. Turning $40^\circ$ further towards west is clockwise: $180^\circ + 40^\circ = 220^\circ$T.
The true bearing $310^\circ$T, written as a compass bearing, is:
$310^\circ$ is $360^\circ - 310^\circ = 50^\circ$ short of north, on the west side: N $50^\circ$ W.
The bearing of $B$ from $A$ is $065^\circ$T. Find the bearing of $A$ from $B$, in degrees.
Back bearing $= 065^\circ + 180^\circ = 245^\circ$T.
Write S $20^\circ$ E as a true bearing, in degrees.
South is $180^\circ$T. Turning $20^\circ$ back towards east is anticlockwise, so subtract: $180^\circ - 20^\circ = 160^\circ$T.
A hiker walks $7$ km on $030^\circ$T, then $4$ km on $120^\circ$T. How far is the hiker from the start, to two decimal places?
Back bearing of leg 1 is $210^\circ$T, so the angle at the turn is $210^\circ - 120^\circ = 90^\circ$. Distance $= \sqrt{7^2 + 4^2} = \sqrt{65} \approx 8.06$ km.
For the hiker in the previous question, find the bearing of the finish from the start, to the nearest degree.
At the start, $\tan^{-1}\left(\dfrac{4}{7}\right) \approx 29.7^\circ$ clockwise of the first leg. Bearing $\approx 030^\circ + 29.7^\circ \approx 060^\circ$T.
A ferry ends $10$ km from its start on a bearing of $127^\circ$T. On what bearing must it sail to go straight back?
Going back is the back bearing: $127^\circ + 180^\circ = 307^\circ$T.
Exam style: A yacht sails $12$ km due north from Port Hacking, then $9$ km due west. Find the bearing of the yacht from Port Hacking, to the nearest degree.
The legs meet at a right angle. At the start, the angle west of north is $\tan^{-1}\left(\dfrac{9}{12}\right) \approx 36.9^\circ$. West of north means subtract from $360^\circ$: $360^\circ - 36.9^\circ \approx 323^\circ$T. (The distance is $\sqrt{12^2 + 9^2} = 15$ km.)
Exam style: A boat sails on a bearing of $060^\circ$T to a buoy, then turns onto a bearing of $170^\circ$T. Find the size of the angle inside the triangle at the buoy, between the two legs, in degrees.
At the buoy, the direction back to the start is the back bearing $060^\circ + 180^\circ = 240^\circ$T. The new leg is $170^\circ$T. Angle $= 240^\circ - 170^\circ = 70^\circ$.