Bearings questions with answers

Questions

1. Question 1Read a bearing

   The diagram shows the positions of $A$ and $B$. Write down the bearing of $B$ from $A$.

   Show the worked solution

   1. Step 1: A bearing is measured clockwise from northThe angle from north, clockwise, to $AB$ is $129^\circ$.
   2. Step 2: Write it with three figuresBearing of $B$ from $A$ $= 129^\circ$
2. Question 2The bearing back

   The bearing of $B$ from $A$ is $081^\circ$. Find the bearing of $A$ from $B$.

   Show the worked solution

   1. Step 1: The north lines at A and B are parallelAngle from south at $B$ to $BA$ $= 81^\circ$ (alt. ∠s)
   2. Step 2: Bearing of A from B: the clockwise angle from north at BBearing of $A$ from $B$ $= 180^\circ + 81^\circ$$= 261^\circ$
3. Question 3How far north or south

   $B$ is $6$ km from $A$ on a bearing of $057^\circ$. Calculate how far $B$ is north of $A$, correct to 3 significant figures.

   Show the worked solution

   1. Step 1: The bearing is the angle from north: which ratio gives the north–south distance y?$\cos 57^\circ = \dfrac{y}{6}$
   2. Step 2: Make y the subject$y = 6 \cos 57^\circ$$y = 3.2678\ldots$$\phantom{y} = 3.27$ km (3 s.f.)
4. Question 4Bearing from distances east and north

   $B$ is $9$ km east and $12$ km south of $A$. Find the bearing of $B$ from $A$, correct to 1 decimal place.

   Show the worked solution

   1. Step 1: The angle α between AB and the north–south line: which ratio?$\tan \alpha = \dfrac{9}{12}$$\alpha = \tan^{-1}\left(\dfrac{9}{12}\right)$$\alpha = 36.8698\ldots^\circ$
   2. Step 2: Turn it into a bearing: a clockwise angle from northBearing $= 180^\circ - 36.8698\ldots^\circ$$= 143.1^\circ$ (1 d.p.)
5. Question 5Two legs at right angles

   A ship sails $5$ km from $A$ to $B$ on a bearing of $140^\circ$, then $3$ km from $B$ to $C$ on a bearing of $230^\circ$. Find the bearing of $C$ from $A$, correct to 1 decimal place.

   Show the worked solution

   1. Step 1: Find ∠ABC from the two bearings at BBearing of $A$ from $B$ $= 320^\circ$ and of $C$ from $B$ $= 230^\circ$, so $\angle ABC = 90^\circ$
   2. Step 2: Which ratio finds ∠BAC?$\tan \angle BAC = \dfrac{3}{5}$$\angle BAC = \tan^{-1}\left(\dfrac{3}{5}\right)$$\angle BAC = 30.9637\ldots^\circ$
   3. Step 3: Use the bearing of B and ∠BAC to find the bearing of CBearing of $C$ from $A$ $= 140^\circ + 30.9637\ldots^\circ$$= 171.0^\circ$ (1 d.p.)
6. Question 6Distance between two places: cosine rule

   From $A$, $B$ is $10$ km away on a bearing of $070^\circ$ and $C$ is $9$ km away on a bearing of $145^\circ$. Find the distance $BC$, correct to 3 significant figures.

   Show the worked solution

   1. Step 1: The angle between the two directions$\angle BAC = 145^\circ - 070^\circ$$= 75^\circ$
   2. Step 2: Which rule finds x?$\text{Cosine rule}$
   3. Step 3: Write the rule with the values, then work it out$x^2 = 10^2 + 9^2 - 2(10)(9) \cos 75^\circ$$x^2 = 134.4125\ldots$$x = 11.5936\ldots$$\phantom{x} = 11.6$ km (3 s.f.)