A direction measured from north. A true bearing is the clockwise angle from north; a compass bearing like N40E rotates from north or south toward east or west.

Pre-Calculus Triangle trigonometry

Bearings and directional applications

Q: How do you convert N40E to a true bearing?

Measure clockwise from north: N40E is a true bearing of 040 degrees.

20 practice questions 0 video lessons Theory + worked examples

Bearings and Triangle Applications

California Pre-Calculus • Standard G-SRT.11 • Triangle Trigonometry

Bearings and Triangle Applications is a topic in Triangle Trigonometry in the California Common Core State Standards. It is aligned to Standard G-SRT.11, which requires students to apply trigonometry to solve problems involving general triangles.

A bearing is a direction measured clockwise from north; navigation problems turn bearings into triangles solved with the Law of Sines or the Law of Cosines.

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Theory

A bearing describes a direction. Two conventions appear:

Compass bearing like \(\text{N}40^\circ\text{E}\): start at north or south, then rotate the stated angle toward east or west.
True bearing like \(040^\circ\): a single angle measured clockwise from north, from \(000^\circ\) to \(360^\circ\).

Navigation and surveying problems turn these directions into triangles: the legs of a journey become sides, and the difference of bearings gives an interior angle. Then the Law of Sines or Law of Cosines finishes the job.

Draw the picture first. Sketch north at each turning point and mark the bearings before you set up any equation.

\(\text{N}40^\circ\text{E}\): \(40^\circ\) clockwise from north, i.e. a true bearing of \(040^\circ\).

A journey's legs and the angle from the bearings form a solvable triangle.

Bearings feed the triangle laws:

\[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C},\qquad c^2=a^2+b^2-2ab\cos C\]

navigation triangles are solved with the Law of Sines and the Law of Cosines

The interior angle usually comes from subtracting bearings (and sometimes using \(180^\circ-\) that difference).

How to solve a bearing problem

Sketch the path, drawing a north line at each turn.
Find the interior angle of the triangle from the bearings.
Choose the law: Cosines for SAS/SSS, Sines otherwise.
Solve for the required distance or direction, and convert back to a bearing if needed.

Example 1 — Read a bearing

A ship sails on a bearing of \(\text{N}40^\circ\text{E}\). What does that mean?

Solution

A compass bearing \(\text{N}40^\circ\text{E}\) is measured 40° clockwise from due north, toward the east.

As a true bearing (clockwise from north, 000–360°), this is \(040^\circ\).

N40E means 40 degrees east of north, a true bearing of 040

Example 2 — Distance with the Law of Cosines

A plane flies \(120\ \text{mi}\) on bearing \(050^\circ\), then \(90\ \text{mi}\) on bearing \(140^\circ\). How far is it from the start?

Solution

The two bearings differ by \(140^\circ-50^\circ=90^\circ\), so the turn angle inside the triangle is \(180^\circ-90^\circ=90^\circ\). Use the Law of Cosines (here just Pythagoras).

\(d^2\)	\(=\)	\(120^2+90^2-2(120)(90)\cos 90^\circ\)
	\(=\)	\(14400+8100=22500\)
\(d\)	\(=\)	\(150\ \text{mi}\)

the plane is 150 miles from the start

Example 3 — Use the Law of Sines for a heading

In that triangle, find the angle at the start between the first leg and the straight-line return path.

Solution

With the \(90^\circ\) angle opposite \(d=150\) and the \(90\)-mi leg opposite the start angle \(\theta\):

\(\sin\theta\)	\(=\)	\(\dfrac{90\sin 90^\circ}{150}=0.6\)
\(\theta\)	\(\approx\)	\(36.9^\circ\)

the angle at the start is about 36.9 degrees

Example 4 — Two observers

Two lookouts \(2\ \text{mi}\) apart sight a fire; the angles from the baseline are \(65^\circ\) and \(75^\circ\). How far is the fire from the first lookout?

Solution

The third angle is \(180^\circ-65^\circ-75^\circ=40^\circ\); use the Law of Sines.

\(d\)	\(=\)	\(\dfrac{2\sin 75^\circ}{\sin 40^\circ}\)
	\(\approx\)	\(3.01\ \text{mi}\)

the fire is about 3.0 miles from the first lookout

Common pitfalls

True bearings are three digits, clockwise from north. Write \(040^\circ\), not \(40^\circ\) from the east.

The interior angle isn't the bearing itself. It comes from the difference of bearings — draw it to be sure.

Convert answers back to bearings. An angle inside the triangle still needs translating into a compass or true bearing.

Frequently asked questions

What is a bearing?

A direction measured from north. A true bearing is the clockwise angle from north (000–360°); a compass bearing like N40°E rotates from north or south toward east or west.

How do you convert N40E to a true bearing?

Measure clockwise from north: \(\text{N}40^\circ\text{E}\) is \(040^\circ\).

How do bearings become a triangle?

Each straight leg of a trip is a side; the angle where legs meet comes from the difference of their bearings. Then apply the Law of Sines or Cosines.

Which law should I use in a navigation problem?

Law of Cosines when you have two legs and the included angle (or three sides); Law of Sines when you have an angle opposite a known side.

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