Three-Figure Bearings

Steps

1. 1

   Stand at the point you are going from and draw a north line.

2. 2

   Turn clockwise from north to the line you want.

3. 3

   Write the angle with three figures: 5° is 005°, 60° is 060°.

Worked example

1. 1

   $B\text{ is past south}$ — B is on the left of the north–south line, so the clockwise angle is more than 180°. A semicircle protractor cannot measure that in one go.

2. 2

   $\text{anticlockwise from north}=123^{\circ }$ — Measure the smaller angle instead: centre on A, zero on north, turning the other way to the line AB.

3. 3

   $360^{\circ }-123^{\circ }=237^{\circ }$ — The clockwise and anticlockwise angles together make one full turn of 360°, so take the small one away.

4. 4

   $\text{bearing of }B\text{ from }A=237^{\circ }$ — It already has three figures, so no zero is needed in front.

Worked example

1. 1

   $30÷5=6\text{ cm}$ — Change the real distance to a length on paper. Each 1 cm stands for 5 km, so divide 30 km by 5.

2. 2

   $\text{draw a north line at }B$ — The bearing is “from B”, so the angle is measured at B. The north line starts there.

3. 3

   $\text{mark }110^{\circ }\text{ clockwise}$ — Centre the protractor on B with zero on the north line. Turn clockwise, because bearings always do: 110° is just past east.

4. 4

   $\text{rule a line from }B\text{ through the mark}$ — Make it longer than 6 cm, so there is room to measure along it.

5. 5

   $\text{measure }6\text{ cm, mark }C$ — Measure 6 cm from B along the line and put a cross. That point is C.

Worked example

1. 1

   $\text{at }P:\text{north line, then }050^{\circ }$ — Start at P, because the first bearing is “from P”. Draw its north line and measure 050° clockwise.

2. 2

   $\text{rule a long line from }P$ — You do not know how far away T is, so make the line long. T is somewhere on it.

3. 3

   $\text{at }Q:\text{north line, then }290^{\circ }$ — Now do the same at Q. 290° is over 180°, so measure 360° − 290° = 70° anticlockwise from north.

4. 4

   $\text{rule a long line from }Q$ — T is also somewhere on this second line.

5. 5

   $\text{mark }T\text{ where the lines cross}$ — T must be on both lines, and they meet at only one point. Put a cross there and label it T.

Worked example

1. 1

   $5÷2=2.5\text{ cm}$ — Change 5 km to a length on paper. Each centimetre stands for 2 km.

2. 2

   $\text{draw the }140^{\circ }\text{ line from }A$ — North line at A, measure 140° clockwise, then rule a long line. T is on it.

3. 3

   $\text{compasses at }B,\text{radius }2.5\text{ cm}$ — Every point 5 km from B lies on a circle around B, so draw an arc of radius 2.5 cm across the line.

4. 4

   $\text{mark }T\text{ where the arc cuts the line}$ — T is on the line and on the arc. If the arc cuts the line twice, use the other facts in the question to pick one.

Worked example

1. 1

   $B\text{ is due east of }A:090^{\circ }$ — Start from a direction you know. Due east is a quarter turn clockwise from north.

2. 2

   $\text{angle }BAC=40^{\circ }\text{, clockwise from }AB$ — C is 40° further round in the clockwise direction, the same way bearings turn.

3. 3

   $090^{\circ }+40^{\circ }=130^{\circ }$ — Turning further clockwise makes the bearing bigger, so add. The bearing of C from A is 130°.

Worked example

1. 1

   $\text{bearing of }B\text{ from }A=030^{\circ }$ — The known direction to start from.

2. 2

   $030^{\circ }-70^{\circ }=-40^{\circ }$ — C is 70° anticlockwise, against the way bearings turn, so take 70° away.

3. 3

   $-40^{\circ }+360^{\circ }=320^{\circ }$ — A bearing cannot be negative. −40° means 40° anticlockwise from north, and adding a full turn of 360° gives the same direction measured clockwise.

4. 4

   $\text{bearing of }C\text{ from }A=320^{\circ }$ — Check: 320° points north-west, just left of north, which matches a line 40° anticlockwise from north.