C7.2; E7.2 - Trigonometry with bearings

C7.2; E7.2 - Trigonometry with bearings

A boat travels 12 km north and then 9 km east. What bearing should it take to return directly to its starting point? C7.2 and E7.2 use right triangles to connect journey lengths, acute angles and the clockwise bearings used on a map.

Bearing diagrams

Start by marking the point named after “from”. Draw north there, with east to the right. A bearing is measured clockwise from north; 090∘090^\circ is east, 180∘180^\circ south and 270∘270^\circ west.

North–south and east–west lines are perpendicular. A journey along one of each makes two legs of a right triangle. Its hypotenuse is the direct connection from the start to the finish, not the total distance travelled.

[DIAGRAM: asset_name: A journey and its direct route; asset_slug: bearing_journey_triangle; caption: The 12 km north and 9 km east legs make a right triangle, with the direct route joining A and C.; description: Exact map with A at the origin, B 12 units north and C 9 units east of B. North arrows at A and C are parallel. AB is labelled 12 km, BC 9 km, angle ABC is marked right, and AC is highlighted. The acute angle between north at A and AC is labelled alpha. A dashed north–south line through C supports the return-bearing interpretation.]
The 12 km north and 9 km east legs make a right triangle, with the direct route joining A and C.

Worked example 1: identifying the direct route

Question. A traveller goes 8 km east from PP to QQ, then 6 km north to RR. Identify the hypotenuse of the journey triangle and write a ratio for the acute angle α\alpha between north at PP and PRPR.

Plan. Sketch the perpendicular legs and compare PRPR with a north line through PP.

The right angle is at QQ, so PRPR is the hypotenuse. To use an angle from north at PP, draw an auxiliary point HH due north of PP and due west of RR. In right triangle PHRPHR, PH=6 kmPH=6\,\mathrm{km} and HR=8 kmHR=8\,\mathrm{km}. Relative to α\alpha, the horizontal change 8 km is opposite and the vertical change 6 km is adjacent.

tan⁡α=86.\tan\alpha=\frac{8}{6}.

Answer. Hypotenuse PRPR, with tan⁡α=8/6\tan\alpha=8/6.

Check. The route is north-east, so α\alpha lies between 0∘0^\circ and 90∘90^\circ. Its eastward change exceeds its northward change, so it lies closer to east than north.

Now reflect only the eastward leg to the west. The length triangle is unchanged, but the clockwise bearing is different.

Worked example 2: moving west instead

Question. A traveller goes 8 km west from UU to VV, then 6 km north to WW. Identify the hypotenuse and write a ratio for the small angle β\beta west of north from UU to WW.

Plan. Use the same perpendicular changes, keeping their directions on the sketch.

Draw ZZ due north of UU and due east of WW. The auxiliary right triangle UZWUZW has UZ=6 kmUZ=6\,\mathrm{km} and ZW=8 kmZW=8\,\mathrm{km}, so its angle at UU is the small angle west of north.

UW=hypotenuseopposite the right angle at Vtan⁡β=86westward over northward change\begin{aligned} UW&=\text{hypotenuse} &&\text{opposite the right angle at }V\\ \tan\beta&=\frac{8}{6} &&\text{westward over northward change} \end{aligned}

Answer. Hypotenuse UWUW, with tan⁡β=8/6\tan\beta=8/6. The bearing is 360∘−β360^\circ-\beta.

Check. WW is north-west of UU, so its bearing must lie between 270∘270^\circ and 360∘360^\circ, unlike the first route.

A route on a stated bearing can also make a right triangle: draw a perpendicular east–west or north–south connection from its endpoint. This supplies the relevant horizontal or vertical component.

Watch out

The sum 8+6=148+6=14 km is the two-leg journey length, not the direct distance. The direct route is the triangle's hypotenuse.

Identify the triangle before doing any trigonometry.

Lengths and bearings

When a direct route and its bearing are known, its north–south and east–west components are the two shorter sides. Use an acute angle inside the triangle. Mark the compass directions separately so positive distances do not lose their meaning.

Worked example 3: a north-east route

Question. A boat travels 24 km from AA to BB on a bearing of 055∘055^\circ. Calculate how far east and how far north BB lies from AA.

Plan. Treat 24 km as the hypotenuse and use the 55∘55^\circ angle from north.

east distance=24sin⁡55∘=19.659649… kmnorth distance=24cos⁡55∘=13.765834… km\begin{aligned} \text{east distance}&=24\sin55^\circ=19.659649\ldots\,\mathrm{km}\\ \text{north distance}&=24\cos55^\circ=13.765834\ldots\,\mathrm{km} \end{aligned}

Sine gives the side opposite the north-based angle; cosine gives the adjacent side.

Answer. BB lies 19.7 km east and 13.8 km north, to three significant figures.

Check. Using unrounded components, their squared sum is 242=57624^2=576. The bearing is closer to east than north, matching the larger eastward component.

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