Bearings and Trigonometry

Worked example

1. 1

   $\text{hypotenuse}=15,\text{angle}=40^{\circ }$ — Sketch it. Draw the north line at $A$, then the path 40° clockwise from it. Draw a line from $B$ back to the north line to make a right angle. The path is the longest side, the hypotenuse.

2. 2

   $\sin 40^{\circ }=\frac{\text{east}}{15}$ — The east side does not touch the 40° angle, so it is the opposite side. Opposite and hypotenuse means sine.

3. 3

   $\text{east}=15\sin 40^{\circ }$ — Multiply both sides by 15 to get east alone.

4. 4

   $\text{east}=15\times 0.6428=9.64\text{ km}$ — Use the calculator in degree mode, and give 3 significant figures.

5. 5

   $\cos 40^{\circ }=\frac{\text{north}}{15}$ — The north side touches the 40° angle, so it is the adjacent side. Adjacent and hypotenuse means cosine.

6. 6

   $\text{north}=15\cos 40^{\circ }=11.5\text{ km}$ — Multiply both sides by 15 and use the calculator. 3 significant figures.

7. 7

   $9.64^2+11.5^2\approx 225=15^2$ — Check with Pythagoras: the two sides squared add up to the path squared.

Worked example

1. 1

   $90^{\circ }<140^{\circ }<180^{\circ }$ — 140° is past east and before south, so the path heads east and south.

2. 2

   $180^{\circ }-140^{\circ }=40^{\circ }$ — The angle inside the right triangle is measured from the south line. It is the gap between 140° and 180°.

3. 3

   $\text{east}=12\sin 40^{\circ }=7.71\text{ km}$ — East is opposite the 40° angle, so use sine. 12 × 0.6428 = 7.71 to 3 significant figures.

4. 4

   $\text{south}=12\cos 40^{\circ }=9.19\text{ km}$ — South is next to the 40° angle, so use cosine. 12 × 0.7660 = 9.19.

Worked example

1. 1

   $\text{right angle at }A$ — Sketch it: $B$ above $A$, $C$ to the right of $A$. Draw a north line at $B$ as well, because the bearing is measured at $B$.

2. 2

   $\tan \angle ABC=\frac{14}{9}$ — In the triangle, $AC$ (14) is opposite the angle at $B$, and $AB$ (9) is next to it. Opposite and adjacent, with no hypotenuse, means tangent.

3. 3

   $\angle ABC={\tan }^{-1}\left(\frac{14}{9}\right)=57.3^{\circ }$ — Inverse tangent gives the angle inside the triangle at $B$.

4. 4

   $\text{due south from }B=180^{\circ }$ — $A$ is straight down from $B$, so the direction $BA$ has a bearing of 180°. The line $BC$ is 57.3° round from that, back towards east.

5. 5

   $180^{\circ }-57.3^{\circ }=122.7^{\circ }$ — Bearings turn clockwise from north. $C$ is 57.3° before reaching south, so take 57.3° from 180°.

6. 6

   $090^{\circ }<122.7^{\circ }<180^{\circ }$ — Check: $C$ is east and south of $B$, so the bearing should be between 090° and 180°. It is.

Worked example

1. 1

   $\tan \angle ACB=\frac{9}{14}$ — At $C$, $AB$ (9) is opposite the angle and $AC$ (14) is next to it, so tangent again.

2. 2

   $\angle ACB={\tan }^{-1}\left(\frac{9}{14}\right)=32.7^{\circ }$ — Inverse tangent gives the angle at $C$.

3. 3

   $\text{due west from }C=270^{\circ }$ — $A$ is straight to the left of $C$, so the direction $CA$ is 270°. $B$ is further round, towards north.

4. 4

   $270^{\circ }+32.7^{\circ }=302.7^{\circ }$ — Going clockwise from west towards north makes the bearing bigger, so add.

5. 5

   $122.7^{\circ }+180^{\circ }=302.7^{\circ }$ — Check with the back bearing: the reverse direction differs by 180°. Both methods agree.

Worked example

1. 1

   $060^{\circ }+90^{\circ }=150^{\circ }$ — Turning right is turning clockwise, so add 90° to the bearing. The second leg is on a bearing of 150°.

2. 2

   $\angle ABC=90^{\circ }$ — The turn is 90°, so the two legs meet at a right angle. $A$ is the start, $B$ the turn and $C$ the finish. Sketch it, with the north line at both $A$ and $B$.

3. 3

   $A{C}^2=10^2+8^2=164$ — $AC$ is opposite the right angle, so it is the hypotenuse. Pythagoras: square the two short sides and add.

4. 4

   $AC=\sqrt{164}=12.8\text{ km}$ — Take the square root, to 3 significant figures.

5. 5

   $\tan \angle BAC=\frac{8}{10}$ — At $A$, $BC$ (8) is opposite the angle and $AB$ (10) is next to it. Tangent.

6. 6

   $\angle BAC={\tan }^{-1}(0.8)=38.7^{\circ }$ — Inverse tangent gives the angle at $A$ between the first leg and the finish.

7. 7

   $060^{\circ }+38.7^{\circ }=98.7^{\circ }$ — The first leg is on 060°. The finish $C$ is turned 38.7° further clockwise from it, as seen from $A$, so add.