Bearings with Trigonometry

The eastward distance is opposite the angle measured from north.

1. The bearing 052^\circ is the angle at A between north and the path AB.
2. Make a right-angled triangle with the north line through A: the 12 km path is the hypotenuse and the side across from the 52^\circ angle runs east. So the distance east is opposite the 52^\circ angle.
3. Opposite = hypotenuse \times\sin(\text{angle}), so the distance east is 12\sin52^\circ km.
4. 12\cos52^\circ is the distance north, the side next to the 52^\circ angle.

The northward distance is adjacent to the bearing angle.

1. In right-angled triangle AHB, the hypotenuse is AB=15 km and the 28^\circ angle is at A, between north and AB.
2. The distance north, AH, is next to the 28^\circ angle, so use cosine: \cos28^\circ=\dfrac{AH}{15}.
3. AH=15\cos28^\circ=13.24\ldots km.
4. B is 13.2 km north of A, to 1 decimal place.

Use sine to relate the eastward distance to the full distance.

1. East is 090^\circ, so the angle between AB and the east line AH is 90^\circ-65^\circ=25^\circ.
2. In right-angled triangle AHB, AH=6 km is next to the 25^\circ angle and AB is the hypotenuse, so \cos25^\circ=\dfrac{6}{AB}. (This is the same as \sin65^\circ=\dfrac{6}{AB}.)
3. Rearrange: AB=\dfrac{6}{\cos25^\circ}=6.62\ldots km.
4. AB=6.6 km to 1 decimal place.

In triangle AHB, find angle HAB, then work out how far round from north AB points.

1. B is south and west of A, so its bearing is between 180^\circ and 270^\circ.
2. In right-angled triangle AHB, AH=7 km runs west from A and HB=10 km runs south. For angle HAB, HB=10 is opposite and AH=7 is next to it, so \tan(\angle HAB)=\dfrac{10}{7}.
3. \angle HAB=\tan^{-1}\left(\dfrac{10}{7}\right)=55.00\ldots^\circ. This is the angle between west and AB.
4. West is 270^\circ, and AB is 55.00\ldots^\circ round from west towards south, so the bearing is 270^\circ-55.00\ldots^\circ=214.99\ldots^\circ.
5. The bearing of B from A is 215^\circ to the nearest degree.

Use the acute angle between the flight path and the east direction.

1. East is 090^\circ, so a bearing of 138^\circ points 138^\circ-90^\circ=48^\circ below the east line.
2. In right-angled triangle AHB, the angle at A is 48^\circ, the hypotenuse is AB=14 km and the distance east, AH, is next to the angle.
3. AH=14\cos48^\circ=9.36\ldots km.
4. The drone is 9.4 km east of A, to 1 decimal place.

Convert both bearings to interior angles, then use the sine rule.

1. Turn each bearing into an angle inside triangle APB.
2. At A: AB points east (090^\circ) and AP is on 035^\circ, so \angle PAB=90^\circ-35^\circ=55^\circ.
3. At B: BA points west (270^\circ) and BP is on 310^\circ, so \angle PBA=310^\circ-270^\circ=40^\circ.
4. Angles in a triangle add up to 180^\circ: \angle APB=180^\circ-55^\circ-40^\circ=85^\circ.
5. Sine rule, pairing each side with the angle opposite it: AP is opposite 40^\circ and AB=8 is opposite 85^\circ, so \dfrac{AP}{\sin40^\circ}=\dfrac{8}{\sin85^\circ}.
6. AP=\dfrac{8\sin40^\circ}{\sin85^\circ}=5.16\ldots, so AP=5.2 km to 1 decimal place.

Find the angle between the two bearings at the harbour.

1. Both bearings are measured clockwise from the same north line at O, so the angle between the routes is the difference: \angle AOB=105^\circ-20^\circ=85^\circ.
2. Two sides and the angle between them are known, so use the cosine rule: AB^2=OA^2+OB^2-2\times OA\times OB\times\cos AOB.
3. AB^2=7^2+11^2-2\times7\times11\times\cos85^\circ=156.57\ldots
4. AB=\sqrt{156.57\ldots}=12.51\ldots, so the buoys are 12.5 km apart.

Use the north line at B: the bearing of A from B helps you find angle ABC.

1. To use the cosine rule for AC we need the angle at B between BA and BC.
2. The bearing of A from B is the back bearing of 040^\circ: 040^\circ+180^\circ=220^\circ.
3. BC is on 145^\circ, so \angle ABC=220^\circ-145^\circ=75^\circ.
4. The back bearing supplies the angle inside the triangle.

5. Cosine rule: AC^2=AB^2+BC^2-2\times AB\times BC\times\cos ABC.
6. AC^2=14^2+9^2-2\times14\times9\times\cos75^\circ=211.77\ldots
7. AC=\sqrt{211.77\ldots}=14.55\ldots, so AC=14.6 km to 1 decimal place.

Find the angle between the route and the south direction.

1. South is 180^\circ, so a bearing of 248^\circ points 248^\circ-180^\circ=68^\circ round from south towards west.
2. In right-angled triangle AHB, the angle at A is 68^\circ, the hypotenuse is AB=18 km and the distance south, AH, is next to the angle.
3. AH=18\cos68^\circ=6.74\ldots km.
4. The boat is 6.7 km south of A, to 1 decimal place.

Find the angle between east and AB first, then measure from north instead.

1. In right-angled triangle AHB, BH=5 km is opposite the angle at A and AH=8 km is next to it, so \tan(\angle HAB)=\dfrac58.
2. \angle HAB=\tan^{-1}\left(\dfrac58\right)=32.00\ldots^\circ. This is the angle between east and AB.
3. A bearing is measured clockwise from north, and east is 090^\circ, so the bearing is 90^\circ-32.00\ldots^\circ=57.99\ldots^\circ.
4. The bearing of B from A is 058.0^\circ to 1 decimal place.

The smaller angle between the routes passes through north.

1. A is on 345^\circ, which is 360^\circ-345^\circ=15^\circ west of north. B is on 055^\circ, which is 55^\circ east of north.
2. So the angle between the routes is \angle AOB=15^\circ+55^\circ=70^\circ.
3. Two sides and the angle between them are known, so use the cosine rule: AB^2=OA^2+OB^2-2\times OA\times OB\times\cos AOB.
4. AB^2=9^2+12^2-2\times9\times12\times\cos70^\circ=151.12\ldots
5. AB=\sqrt{151.12\ldots}=12.29\ldots, so AB=12.3 km to 1 decimal place.

At the northern station, compare its bearing to the south direction.

1. At A: AB points north and AP is on 070^\circ, so \angle BAP=70^\circ.
2. At B: BA points south (180^\circ) and BP is on 135^\circ, so \angle ABP=180^\circ-135^\circ=45^\circ.
3. Angles in a triangle: \angle APB=180^\circ-70^\circ-45^\circ=65^\circ.
4. Sine rule: AP is opposite the 45^\circ angle and AB=10 is opposite the 65^\circ angle, so \dfrac{AP}{\sin45^\circ}=\dfrac{10}{\sin65^\circ}.
5. AP=\dfrac{10\sin45^\circ}{\sin65^\circ}=7.80\ldots, so AP=7.8 km to 1 decimal place.

Show that the two legs are perpendicular before finding the angle at the start.

1. At B, the bearing of A is the back bearing of 125^\circ: 125^\circ+180^\circ=305^\circ.
2. From BA on 305^\circ, turning clockwise past north to BC on 035^\circ is (360^\circ-305^\circ)+35^\circ=90^\circ. So \angle ABC=90^\circ.
3. In right-angled triangle ABC, \tan(\angle BAC)=\dfrac{BC}{AB}=\dfrac{12}{9}, so \angle BAC=53.13\ldots^\circ.
4. From A, C is 53.13\ldots^\circ anticlockwise from AB (nearer north), so the bearing of C from A is 125^\circ-53.13\ldots^\circ=71.86\ldots^\circ.
5. The right angle gives the triangle angle; then reverse the bearing of AC.

6. The bearing of A from C is the back bearing: 71.86\ldots^\circ+180^\circ=251.86\ldots^\circ, which is 251.9^\circ to 1 decimal place.

Find a distance to the boat using the sine rule, then resolve it perpendicular to the coast.

1. At A: AB points east and AP is on 030^\circ, so \angle PAB=90^\circ-30^\circ=60^\circ. At B: BA points west (270^\circ), so \angle PBA=305^\circ-270^\circ=35^\circ.
2. So \angle APB=180^\circ-60^\circ-35^\circ=85^\circ.
3. Sine rule: AP is opposite the 35^\circ angle and AB=10 is opposite the 85^\circ angle, so AP=\dfrac{10\sin35^\circ}{\sin85^\circ}=5.757\ldots km.
4. The distance from the coastline is the perpendicular PH. In right-angled triangle AHP, PH is opposite the 60^\circ angle and AP is the hypotenuse, so PH=AP\sin60^\circ.
5. PH=5.757\ldots\times\sin60^\circ=4.98\ldots km, so the boat is 5.0 km from the coastline, to 1 decimal place.

Use the smaller angle between the two routes in the triangle area formula.

1. A is on 330^\circ, which is 360^\circ-330^\circ=30^\circ west of north. B is on 070^\circ, which is 70^\circ east of north.
2. So the angle between OA and OB is 30^\circ+70^\circ=100^\circ.
3. For two sides and the angle between them, area =\dfrac12ab\sin C.
4. Area =\dfrac12\times8\times13\times\sin100^\circ=51.21\ldots
5. The area of triangle OAB is 51.2 km^2 to 1 decimal place.

At B, use the bearing of A from B to find the angle inside the triangle.

1. At A: AB is on 040^\circ and AP is on 100^\circ, so \angle BAP=100^\circ-40^\circ=60^\circ.
2. At B: the bearing of A from B is the back bearing 040^\circ+180^\circ=220^\circ. BP is on 155^\circ, so \angle ABP=220^\circ-155^\circ=65^\circ.
3. Angles in a triangle: \angle APB=180^\circ-60^\circ-65^\circ=55^\circ.
4. Carry the two bearing differences into the triangle before using the sine rule.

5. Sine rule: BP is opposite the 60^\circ angle and AB=12 is opposite the 55^\circ angle, so \dfrac{BP}{\sin60^\circ}=\dfrac{12}{\sin55^\circ}.
6. BP=\dfrac{12\sin60^\circ}{\sin55^\circ}=12.68\ldots, so BP=12.7 km to 1 decimal place.

Use the cosine rule to find the angle between the eastward baseline and the tower.

1. All three sides are known, so use the cosine rule to find the angle at A: \cos A=\dfrac{AB^2+AC^2-BC^2}{2\times AB\times AC}.
2. \cos(\angle BAC)=\dfrac{10^2+7^2-8^2}{2\times10\times7}=\dfrac{85}{140}
3. \angle BAC=\cos^{-1}\left(\dfrac{85}{140}\right)=52.61\ldots^\circ. This is the angle between east (along AB) and AC.
4. C is north of AB and east is 090^\circ, so the bearing of C from A is 90^\circ-52.61\ldots^\circ=37.38\ldots^\circ.
5. The bearing is 037.4^\circ to 1 decimal place.

The right angle at the common origin lets you find the missing side first.

1. \angle AOB=105^\circ-15^\circ=90^\circ, so triangle AOB is right-angled at O, with hypotenuse AB=17 km.
2. Pythagoras: OB=\sqrt{17^2-8^2}=\sqrt{225}=15 km.
3. In the right-angled triangle, \tan(\angle OAB)=\dfrac{OB}{OA}=\dfrac{15}{8}, so \angle OAB=61.92\ldots^\circ.
4. From A, the bearing of O is the back bearing of 015^\circ: 015^\circ+180^\circ=195^\circ.
5. Carry the derived side and angle into the final bearing calculation.

6. B is 61.92\ldots^\circ anticlockwise from AO, so the bearing of B from A is 195^\circ-61.92\ldots^\circ=133.07\ldots^\circ, which is 133.1^\circ to 1 decimal place.

Use the sine rule to find a sloping side, then use an area formula.

1. At A: \angle PAB=90^\circ-60^\circ=30^\circ. At B: BA points west (270^\circ), so \angle PBA=335^\circ-270^\circ=65^\circ.
2. So \angle APB=180^\circ-30^\circ-65^\circ=85^\circ.
3. Sine rule: AP is opposite the 65^\circ angle and AB=8 is opposite the 85^\circ angle, so AP=\dfrac{8\sin65^\circ}{\sin85^\circ}=7.278\ldots km.
4. Now use area =\dfrac12ab\sin C with sides AB and AP and the 30^\circ angle between them.
5. Area =\dfrac12\times8\times7.278\ldots\times\sin30^\circ=14.55\ldots, so the area is 14.6 km^2 to 1 decimal place.

Find the distance between the vessels, then find the angle at the first vessel.

1. A is on 285^\circ, which is 360^\circ-285^\circ=75^\circ west of north. B is 35^\circ east of north. So \angle AOB=75^\circ+35^\circ=110^\circ.
2. Cosine rule: AB^2=11^2+16^2-2\times11\times16\times\cos110^\circ=497.39\ldots, so AB=22.30\ldots km.
3. Cosine rule again for the angle at A: \cos(\angle OAB)=\dfrac{OA^2+AB^2-OB^2}{2\times OA\times AB}=\dfrac{11^2+22.30\ldots^2-16^2}{2\times11\times22.30\ldots}
4. \angle OAB=42.38\ldots^\circ.
5. From A, the bearing of O is the back bearing: 285^\circ-180^\circ=105^\circ.
6. Use the derived angle at A to turn from the back bearing towards north.

7. B is 42.38\ldots^\circ anticlockwise from AO (nearer north), so the bearing of B from A is 105^\circ-42.38\ldots^\circ=62.61\ldots^\circ, which is 062.6^\circ to 1 decimal place.