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Lesson 7, Topic 2
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# Bearings and Distances

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Bearings are the clockwise angular relationship between two distant places, measured in degrees from the north.

The relevant directions are the four cardinal points; North, South, East and West.

For example, a bearing of N32°E means a turn of 32° from the north towards the direction of the East.

Note: In sketching the diagram representing bearing problems, the term due North, South, East, or West means on the straight line of any cardinal point mentioned.

### Example 1

A village P is 10km from a point X on a bearing 025° from X.

Another village Q, is 6km from X on a bearing of 162°. Calculate the distance and bearing of P from Q.

Solution:

$$\scriptsize \hat{X} = 162^o \: – \: 25^o = 137^o$$

Apply Cosine rule to find |PQ|

c2 = a2 + b2 – 2abCosC

|PQ|2 = |XP|2 + |XQ|2 – 2|XP||XQ| Cos X

|PQ|2 = $$\scriptsize 10^2 \: + \: 6^2 \: – \: 2 \: \times \: 10 \: \times \: 6 \: \times \: Cos 137^o$$

|PQ|2 = $$\scriptsize 100 \: + \: 36 \: \: + \: 87.768$$

|PQ|2 = $$\scriptsize 233.768$$

|PQ| = $$\scriptsize \sqrt{233.768}$$

|PQ| = $$\scriptsize 14.9588$$

|PQ| $$\scriptsize \cong 14.96 \: km$$

Using Sine rule we can find <Q

$$\frac{SinQ}{10} = \frac{Sin137}{14.96}$$

$$\scriptsize Sin Q = \normalsize \frac{Sin137 \: \times \: 10}{14.96}$$

$$\scriptsize Sin Q = \normalsize \frac{0.681998 \: \times \: 10}{14.96}$$

$$\scriptsize Sin Q = \normalsize \frac{6.81998}{14.96}$$

$$\scriptsize Sin Q = 0.45588$$

Q = Sin-1(0.45588)= 27.12°

$$\scriptsize \alpha$$ = bearing of P from Q = 27.12 – 18 = 9.12°

### Example 2

2. Two cyclists X and Y leave town at the same time. Cyclist X travels at the rate of 5km/h on a bearing of 049° and cyclist Y travels at the rate of 9km/h on a bearing of 319°.

After travelling for two hours, Calculate correct to the nearest whole number, the:

i. distance between cyclist X and Y

ii. bearing of cyclist X from Y.

Solution:

Speed = $$\frac{distance}{times}$$

Therefore, 5km/h = $$\frac{distance}{2hrs}$$

D=10km=|QX|

9km/h = $$\frac{distance}{2hrs}$$

D= 18km= |QY|

i. |XY|2= 182+102

|XY| = $$\scriptsize \sqrt{424}$$

|XY| = 20.5913km

ii. tan Y = $$\frac{10}{18}$$

Y = $$\scriptsize tan^{-1} \left(\normalsize \frac{10}{18} \right)$$

Y = 29.06°

Bearing of X from Y = 180° – (41 + 29.06°)

= 109.94°

≅ 110°

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