Bearings with Sine and Cosine Rules

Steps

1. 1

   Sketch it, with a north line at every point.

2. 2

   Find the angle inside the triangle from the bearings.

3. 3

   Cosine rule for a distance, sine rule for an angle.

4. 4

   Turn the angle back into a bearing.

Worked example

1. 1

   $060^{\circ }+180^{\circ }=240^{\circ }$ — We need the angle at $Q$, so we need both directions from $Q$. The way back from $Q$ to $P$ is the back bearing: add 180°, a half turn.

2. 2

   $\angle PQR=240^{\circ }-170^{\circ }$ — Both directions now start at $Q$ and are measured from the same north line. The angle between them is the difference.

3. 3

   $\angle PQR=70^{\circ }$ — That is the angle inside the triangle at $Q$, between the two legs.

Worked example

1. 1

   $P{R}^2=8^2+5^2-2(8)(5)\cos 70^{\circ }$ — We know two sides (8 and 5) and the angle between them, and want the side opposite that angle. That is the cosine rule.

2. 2

   $P{R}^2=64+25-80\cos 70^{\circ }$ — Square the sides: 64 and 25. Multiply 2 × 8 × 5 = 80.

3. 3

   $P{R}^2=89-27.36=61.64$ — Work out $80\cos 70^{\circ }=27.36$ first, then subtract it from 89.

4. 4

   $PR=\sqrt{61.64}=7.851$ — Square root to get $PR$. Keep all 4 figures, because the next part uses them.

5. 5

   $PR=7.85\text{ km}$ — Round to 3 significant figures for the answer.

Worked example

1. 1

   $\frac{\sin P}{5}=\frac{\sin 70^{\circ }}{7.851}$ — Sine rule, with the unknown angle on top. The side 5 faces angle $P$, and 7.851 faces the 70° angle, so each is a side with its facing angle.

2. 2

   $\sin P=\frac{5\times \sin 70^{\circ }}{7.851}=0.5985$ — Multiply both sides by 5, then work it out.

3. 3

   $P={\sin }^{-1}(0.5985)=36.8^{\circ }$ — Inverse sine gives the angle inside the triangle. It is acute, because 5 is the shorter side and so $P$ is smaller than 70°.

4. 4

   $060^{\circ }+36.8^{\circ }=096.8^{\circ }$ — The second leg turns clockwise (170° is bigger than 060°), so $R$ is clockwise from the first direction. Add the angle, and write three figures.

Worked example

1. 1

   $285^{\circ }-180^{\circ }=105^{\circ }$ — The back bearing, $Y$ to $X$. The bearing is over 180°, so take 180° away. Adding would go past 360°.

2. 2

   $\angle XYZ=230^{\circ }-105^{\circ }=125^{\circ }$ — Both directions start at $Y$, so the angle between them is the difference.

3. 3

   $X{Z}^2=6^2+9^2-2(6)(9)\cos 125^{\circ }$ — Two sides (6 and 9) and the angle between them: cosine rule for $XZ$.

4. 4

   $\cos 125^{\circ }=-0.5736$ — 125° is obtuse, so its cosine is negative. Keep the minus sign.

5. 5

   $X{Z}^2=117-108(-0.5736)=117+61.95=178.95$ — Square the sides: 36 + 81 = 117. Minus times minus is plus, so 61.95 is added.

6. 6

   $XZ=\sqrt{178.95}=13.38$ — Square root. Keep 4 figures for the next step.

7. 7

   $\frac{\sin X}{9}=\frac{\sin 125^{\circ }}{13.38}$ — Sine rule. The side 9 ($YZ$) faces angle $X$, and 13.38 faces the 125° angle.

8. 8

   $X={\sin }^{-1}\left(\frac{9\sin 125^{\circ }}{13.38}\right)=33.4^{\circ }$ — Multiply by 9 and use the inverse sine. $X$ is acute because 9 is the shorter side.

9. 9

   $285^{\circ }-33.4^{\circ }=251.6^{\circ }$ — The path turned left (230° is smaller than 285°), so $Z$ is anticlockwise from the direction of $Y$. Take the angle away.

Worked example

1. 1

   $\angle CAB=90^{\circ }-47^{\circ }=43^{\circ }$ — Due east is 090°, and $AC$ is at 047°. The angle between $AC$ and $AB$ is the gap between them.

2. 2

   $\frac{\sin B}{40}=\frac{\sin 43^{\circ }}{58}$ — Sine rule. The side 58 faces the 43° angle, and 40 faces angle $B$.

3. 3

   $\sin B=\frac{40\sin 43^{\circ }}{58}=0.4703$ — Multiply both sides by 40, then work it out.

4. 4

   $B={\sin }^{-1}(0.4704)=28.1^{\circ }$ — Inverse sine. It is acute, because 40 is shorter than 58, so $B$ is smaller than 43°.

5. 5

   $270^{\circ }+28.1^{\circ }=298.1^{\circ }$ — From $B$, $A$ is due west, which is 270°. $C$ is north of the line, so it is 28.1° further clockwise, towards north. Add.