Bearings are used in order to specify a direction relative to north. The bearing is a three-digit angle measured clockwise from the northern direction. In this example, the bearing of A from O is $045^o$. The bearing of B from O is $135^o$. And finally, the bearing of C from O is $315^o$.

## Applications

### Pythagoras’ theorem

Bearings can be used in conjunction to Pythagoras’ theorem in order to solve application-typed problems.

A hiker travels 3km due north and then 4km due east from point A to point B.
a) What is the bearing of B from A?

b) What is the bearing of A from B?

c) What is the distance between A and B?

Using this information, the following diagram can be deduced: Trigonometric ratios is used to calculate the angles inside the triangle. $tan (A) = \frac{4}{3}$ $A = 53.13$ $tan (B) = \frac{3}{4}$ $B = 36.87$

However, bearings are calculated clockwise from the northern direction. Therefore, the bearing of B from A is $053^o$. The bearing of A from B is $233^o$.

A review of Pythagoras’ theorem states that $a^2 + b^2 = c^2$. In this case, a = 3, = 4, and c is the distance AB. Solving this, $c = \sqrt{3^2+4^2}$ $c = 5$ km

Point A is on a bearing of 233 degrees, 5km from point B.

### Cosine and sine rules

The sine and cosine rules can also be used alongside bearings in order to solve a wider variety of problems.

A hiker leaves base camp (B) and travels on a bearing of 135° for 4km to find a lake (L). She then travels on a bearing of 075° for 5km to a cabin (C).
a) Find angle BLC.
b) What is the distance from B to C?
c) What is the bearing of B from C?

Using this information, the following diagram can be deduced: In this diagram, the black lines denote the information provided. The pink marks indicate the information (part a, b, c) requested in the problem.

However, in order to find the angle BLC, the triangle must be split into two right-angled triangles. This is shown below: Since the angles in a triangle add up to 180°,  the unknown angle in the blue triangle is 45°. Therefore, the angle BLC is 45+75° = 120°.

With this angle, the cosine rule can be applied to find the length BC in the first triangle. $BC^2 = 4^2 + 5^2 - 2*4*5cos(120)$ $BC = 7.81$ km

To find the bearing of B from C, the sine rule could be used on the first triangle again. $\frac{7.81}{sin(120)} = \frac{4}{sin(C)}$ $sin(C) = \frac{4sin(120)}{7.81}$ $C = 26.33^o$

To convert that into bearings, it would be 243.7°.\$

Thus, the base camp is on a bearing of 244° compared to the cabin, and is 7.81km away.