Bearings

Bearings are angles measured in a clockwise direction from North. This system is used to describe the direction of a point or object relative to a reference point, typically North.

This north direction is usually provided in the maths exam question. We then measure the required angle in a clockwise direction. All bearings need to be given in three figures, e.g. 008°, 010°, 045°, 080°, 128°, 260°, etc.

For Example:

The diagram shows three points X, Y and Z.

The angles are measured clockwise from the north line. 

• The bearing of X from Z is 045º
• The bearing of Y from Z is 250º

Example 6.2.1:

The points X, Y and Z are located such that Y is 15 km south of X, Z is 20 km from X on a bearing of 270º.

Calculate, correct to:
(a) two significant figures,  |YZ|
(b) the nearest degree, the bearing of Y from Z

Solution

(a)

Using Pythagoras’ Theorem:

⇒ \(\scriptsize |YZ|^2 = 15^2 \: + \: 20^2  \\ \scriptsize = 225 \: + \: 400 \\ \scriptsize = |YZ|^2 = 625 \\\scriptsize = |YZ| = \sqrt{625} \\ \scriptsize|YZ| = 25 \: km \)

Answer ⇒  |YZ| = 25 km

(b)

To the nearest degree, the bearing of Y from Z

Solution

tan θ = \(  \frac{15}{20} \)

θ = \( \scriptsize tan^{-1} \left(\normalsize \frac{15}{20}\right) \)

θ = 26.8699º

Bearing of Y from Z

= 90° + 36.8699

= 126.8699

Answer \( \scriptsize \simeq  127^o\)

Example 6.2.2:

A town Y is 200 km from town X in a direction 040º.

(a) Find the bearing of town Y from town X.
(b) How far is Y East of X?

(WAEC)

Solution:

(a)

Bearing of X from Y is 90º + 90º + 40º = 220º

(b)

Consider right-angle triangle ΔXEY

Considering angle Y = 40º

XE = Opposite = ?

XY = Hypotenuse = 200 km

Recall

\( \scriptsize Sin \: \theta \: = \normalsize \frac{Opp}{Hyp} \)

\( \scriptsize Sin 40^o = \normalsize \frac{\overline{XE}}{200} \)

i.e \( \scriptsize \overline{XE} = 200 \: \times \: Sin 40^o \)

\( \scriptsize \overline{XE} = 200 \: \times \: 0.6427 \)

\( \scriptsize \overline{XE} = 128.5575 \)

\( \scriptsize \overline{XE} \: \approx \: 128.6 \: km \)

Example 6.2.3:

A cottage is on a bearing of 200° and 110° from Dogbe’s and Manu’s farms, respectively. If Dogbe walked 5 km and Manu 3 km from the cottage to their farms,

find:
(i) correct to two significant figures, the distance between the two farms;
(ii) correct to the nearest degree, the bearing of Manu’s farm from Dogbe’s.

Solution

(i)

Let Dogbe = D, Cottage = C and Manu = M

Note: The angles in purple and red are alternate angles

|DM| = the distance between the two farms

From the sketch, ∠DCM = 90º

That means we can use Pythagoras’ theorem to find |DM|

i.e. \( \scriptsize |DM|^2 = |CD|^2 \: + \: |CM|^2 \)

|DM| = \( \scriptsize \sqrt{3^2 \: + \: 5^2} \\ = \scriptsize \sqrt{9 \: + \: 25} \\ = \scriptsize \sqrt{34}\\ = \scriptsize 5.831 \: km \\ \scriptsize \approx 5.8 \: km\)

Answer ⇒  Distance between two farms ≅ 5.8 km

(ii)

tan θ = \( \frac{opp}{adj} \\ = \frac{3}{5} \\ = \scriptsize 0.6\)

θ = \( \scriptsize tan^{-1} \left(0.6 \right)\)

θ = 30.9638º

Bearing = 30.9638° + 200° (see diagram)

= 230.9638°

Answer ⇒  the bearing of Manu’s farm from Dogbe’s ≅ 231º (nearest degree)

Example 6.2.4:

Three ships P, Q, and R are at sea. The bearing of Q from P is 030º and the bearing of P from R is 300º. If |PQ| = 5 km and |PR| = 8 km;

(i) Illustrate the information in a diagram
(ii) Calculate, correct to three significant figures, the:
– distance between Q and R;
– bearing of R from Q

Solution

(i)

(ii)

Calculate, correct to three significant figures, the:

• distance between Q and R (blue line);

Solution

Step 1: Since the ΔQPR is a right-angled triangle, then by Pythagoras

QR² = QP² + PR²

QR² = 5² + 8²

QR² = 25 + 64

QR² = 89

QR = \( \scriptsize \sqrt{89} \)

QR = 9.433

QR ≅ 9.433 km (to 3 significant figures)

• bearing of R from Q

Solution

To calculate the bearing of R from Q

Step 1: From triangle QPR, let \( \scriptsize \hat{PQR} = \theta \)

Step 2: Sin θ = \( \frac{opp}{hyp} \\ = \frac{8}{9.43} \\ = \scriptsize 0.8483 \)

θ = Sin^-1 (0.8483)

θ  = 58º

Hence the bearing of R from Q is 270º – (60 + 58)º

= 270º – 118º

Bearing of R from Q = 152º