Year 10 Core — Directions and Bearings (6D)

Today's lesson

Bearings are how navigators, pilots, surveyors and orienteers describe a direction. By the end of today you should be able to read, write, and apply true bearings in trig word problems.

Learning intentions

Know how a true bearing is measured — clockwise from due north, written with 3 digits and a $°$T
State the bearing of one point from another (and the reverse bearing — they differ by $180°$)
Combine bearings with right-triangle trig to find east/west and north/south distances

Part 1 — What is a true bearing? (~6 min)

The rules for true bearings

Always measured clockwise from due north (north = $000°$T).
Always written with 3 digits — e.g. $048°$T not $48°$T.
Range is $000°$ to $360°$ (or $000°$T equivalently to $360°$T).
The reverse (back) bearing differs by $180°$. If $A$ is on bearing $070°$T from $B$, then $B$ is on bearing $250°$T from $A$.

📺 Walkthrough: the eight cardinal directions and their three-digit true bearings.

Building understanding — write each direction as a true bearing

a) $000°$T b) $045°$T c) $090°$T d) $135°$T
e) $180°$T f) $225°$T g) $270°$T h) $315°$T

Building understanding — opposite (reverse) bearings: add or subtract $180°$

$020°$T
$262°$T
$155°$T
$344°$T

a) $200°$T b) $082°$T (262−180) c) $335°$T (155+180) d) $164°$T (344−180)

Part 2 — Stating a direction from a diagram (Example 7, ~10 min)

When you see an angle drawn at a point relative to N/S/E/W, convert it to a true bearing by starting from north and going clockwise.

EXAMPLE 7 — bearings from a compass diagram

From point $O$, point $A$ is drawn $35°$ above east (i.e. $35°$ measured up from the east axis toward north). Point $B$ is $45°$ below east. Point $C$ is $30°$ below west. Point $D$ is $70°$ above west. Find the true bearing of each point from $O$.

$A$ — start at $090°$T (east), turn $35°$ back toward N (anticlockwise): $A=90°-35°=\boxed{055°}$T.
$B$ — east + $45°$ clockwise toward S: $B=90°+45°=\boxed{135°}$T.
$C$ — west = $270°$T, then $30°$ further clockwise toward S: $C=270°-30°=\boxed{240°}$T.
$D$ — west $270°$, then $70°$ back toward N: $D=270°+70°=\boxed{340°}$T.

Reverse bearings: bearing of $O$ from $A$ = $055°+180°=\boxed{235°}$T. Bearing of $O$ from $D$ = $340°-180°=\boxed{160°}$T.

Part 3 — Using bearings with trig (Example 8, ~14 min)

📺 Walkthrough: a ship sails 5 km due south, then 11 km on bearing 120°T. Find the east and south components.

EXAMPLE 8 — ship on a bearing

A ship travels due south for $5$ km, then on a bearing of $120°$T for $11$ km.
(a) Find how far east the ship is from its starting point (to 2 d.p.).
(b) Find how far south the ship is from its starting point.

Draw the diagram. Bearing $120°$T from the turning point is $30°$ east of south (since $120°-90°=30°$).
Draw the right triangle with the 11 km leg as the hypotenuse, $x$ = east component, $y$ = south component, $30°$ at the turning point measured from south. Then $x$ is opposite $30°$? Actually with $30°$ from the south axis, the east leg is the side opposite 60° (or adjacent to 30° if you measure from south down). Easiest: just use the angle from the south axis.
(a) East component: $\sin 30°=\dfrac{x}{11}$ if $30°$ is measured between south and the path. But Cambridge takes $30°$ as the angle between south-axis and the 11 km leg, so the leg perpendicular to south (i.e. east leg) sits opposite the angle $90°-30°=60°$, which gives $\cos 30°=\dfrac{x}{11}$ ⇒ $x=11\cos 30°\approx \boxed{9.53\text{ km east}}$.
(b) South component of leg 2: $\sin 30°=\dfrac{y}{11}$ ⇒ $y=11\sin 30°=5.5$ km. Total south from start = $5+5.5=\boxed{10.5\text{ km}}$.

Now you try: Ship goes due south 8 km, then bearing $160°$T for 12 km. East = $12\sin 20°\approx 4.10$ km; total south = $8+12\cos 20°\approx 19.28$ km.

Drawing tip. Always sketch a small north arrow at the point where the bearing is measured. Then convert the bearing into an angle inside the right triangle (often $\text{bearing}-90°$, $180°-\text{bearing}$, etc.). The clean way: measure how far the path is from the nearest cardinal direction (N/S/E/W) and put that inside the triangle.

Practice 3.1 — bearings & components

A plane flies $100$ km on a true bearing of $070°$T. How far east and how far north of the start? (2 d.p.)
A hiker walks $6$ km on a bearing of $215°$T. How far west and how far south? (2 d.p.)
A boat sails $4$ km due north, then $5$ km on bearing $050°$T. How far east of the start, and how far north of the start? (2 d.p.)

a) east = $100\sin 70°\approx 93.97$ km; north = $100\cos 70°\approx 34.20$ km
b) west = $6\sin 35°\approx 3.44$ km; south = $6\cos 35°\approx 4.91$ km (215°T is 35° west of south)
c) east = $5\sin 50°\approx 3.83$ km; total north = $4+5\cos 50°\approx 7.21$ km

Part 4 — Quick quiz (5 min)

Working program — Cambridge Ex 6D

Section	Foundation	Standard (set work)	Advanced
Fluency 1–7	1–5	1, 2, 4–6	2, 4, 5, 7
Problem-solving 8–11	8, 9	8–10	10, 11
Reasoning 12–14	12	12, 13	12–14
Enrichment 15–16	—	—	15, 16

Exit ticket — write in your book

Before you pack up, write a one-line answer to each:

What does the bearing $030°$T mean in plain English?
If $B$ is on bearing $112°$T from $A$, what is the bearing of $A$ from $B$?
You walk $20$ m on bearing $090°$T then $20$ m on bearing $180°$T. Sketch the path and state where you are relative to start.