Ships, planes and surveyors rarely meet right triangles. With the law of sines (handling the ambiguous case when it comes up) and the law of cosines, you can solve any triangle. Navigation adds one more idea: directions measured from north.
Key idea
A bearing is measured clockwise from north, from $0^\circ$ to $360^\circ$. It can also be written in compass form such as N $40^\circ$ E: face north, turn $40^\circ$ toward the east. The standard-position angle (counterclockwise from east) is $90^\circ - \text{bearing}$, adjusted by $360^\circ$. To solve: use the law of cosines for two sides and the angle between them (SAS) or three sides (SSS), and the law of sines when you know an angle and the side across from it.
Why it works
Navigators use north because a compass needle points there, and they turn clockwise because that is how compass dials are marked. Math class uses east and counterclockwise. Both describe the same directions; north is $90^\circ$ in standard position, and turning clockwise subtracts.
Worked examples
Worked example: Converting directions
Write N $35^\circ$ W as a bearing and as a standard-position angle.
- Face north and turn $35^\circ$ toward the west. Clockwise from north that is $360^\circ - 35^\circ = 325^\circ$.
- Standard angle: $90^\circ - 325^\circ = -235^\circ$, and adding $360^\circ$ gives $125^\circ$ (up and to the left, as expected).
Answer: Bearing $325^\circ$; standard angle $125^\circ$
Worked example: Two hikers (law of cosines)
Two hikers leave camp. One walks $3$ km on a bearing of $20^\circ$; the other walks $8$ km on a bearing of $80^\circ$. How far apart are they?
- The angle between their paths at camp is $80^\circ - 20^\circ = 60^\circ$.
- $$d^2 = 3^2 + 8^2 - 2(3)(8)\cos 60^\circ = 9 + 64 - 24 = 49.$$
Answer: $d = 7$ km
Worked example: Across a river (law of sines)
Points $A$ and $B$ are $100$ m apart on one bank. A tree $C$ is across the river with $\angle A = 45^\circ$ and $\angle B = 105^\circ$. Find $BC$.
- $\angle C = 180^\circ - 45^\circ - 105^\circ = 30^\circ$. Side $AB = 100$ is across from $C$.
- $BC$ is across from $A$: $\dfrac{BC}{\sin 45^\circ} = \dfrac{100}{\sin 30^\circ}$.
- $BC = \dfrac{100 \cdot \frac{\sqrt{2}}{2}}{\frac{1}{2}} = 100\sqrt{2}$.
Answer: $100\sqrt{2} \approx 141.4$ m
Common mistake
Measuring a bearing from the $x$-axis. A bearing of $40^\circ$ is not $40^\circ$ in standard position; it is $50^\circ$, because it starts at north. Sketch a compass rose with N at the top before you draw any triangle.