A journey in two legs
A ship leaves a harbor A and sails 8 km on a bearing of 040° to a point B. There it changes course and sails 5 km on a bearing of 160° to a point C. How far is it from C straight back to A, and on what bearing must it sail to get there?
The three points make a triangle. Two of its sides, 8 km and 5 km, are known. To use the cosine rule or the sine rule, an angle inside the triangle is needed as well, and the bearings do not give it directly: each bearing is measured from north, not from another side of the triangle.
The journey drawn to scale: 8 km on 040° from A to B, then 5 km on 160° from B to C, with a north line at each point.
The angle at B, from the north line at B
Draw a north line at B. The north lines at A and B are parallel, so the bearing of A from B is the back bearing of the first leg: 040° + 180° = 220°.
Standing at B and turning clockwise from north, the direction back to A is at 220° and the direction on to C is at 160°. The angle ABC between them is 220° − 160° = 60°.
The first leg on its own. The bearing of A from B, measured clockwise from the north line at B, is 040° + 180° = 220°.
The same angle, from the turn
There is a second way to see it. At B the heading changes from 040° to 160°, a turn of 160° − 40° = 120° to the right. If the ship had not turned, it would have carried straight on along the line AB, away from A.
The straight-on direction and the direction back to A make a straight line at B, an angle of 180°. The new leg BC splits that straight angle into the turn, 120°, on one side, and the angle inside the triangle on the other. So the angle inside is 180° − 120° = 60°, as before.
The 120° is the angle the ship turned through, outside the triangle. Using it in place of the inside angle is the usual mistake here.
The cosine rule gives the distance home
Now the triangle has two sides, BA = 8 km and BC = 5 km, and the angle between them, 60°. That is the case for the cosine rule. Call the distance home d:
, so km.
Check against the drawing. At B the second leg sets off only 60° away from the direction home, so it carries the ship partly back toward A, and the ship ends up closer to A than B was: 7 km is less than 8 km. Had the 120° been used, would be 89 + 40 = 129 and km, farther from A than B ever was, which the drawing shows is wrong.
The same triangle, taken out of the map: sides of 8 km and 5 km meet at B at 60°, and the distance home d faces that angle.
The bearing home
To sail home, the ship needs the bearing of A from C. First find the angle at C, between the way back to B and the way to A. All three sides are now known, so use the cosine rule for an angle. The side facing C is AB = 8 km: , so , to 1 decimal place. The cosine is positive, so the angle is acute.
Now draw a north line at C. The bearing of B from C is the back bearing of the second leg: 160° + 180° = 340°. From C, the direction to A is 81.8° round from the direction to B, turning counterclockwise, toward the west, as the drawing shows. So the bearing of A from C is 340° − 81.8° = 258.2°, which is 258° to the nearest degree.
Check through the angle at A. Its cosine is , so the angle is 38.2°. The three angles add up: 60° + 81.8° + 38.2° = 180°. From A, C lies 38.2° clockwise from B, so the bearing of C from A is 040° + 38.2° = 078.2°, and the bearing back is 078.2° + 180° = 258.2°, the same.
The sine rule would give sin C = 8 sin 60° ÷ 7 = 0.990, which fits both 81.8° and 98.2°. The cosine rule gives one angle, and its sign settles which.
A journey for the sine rule
A ship at P sees a lighthouse L on a bearing of 050°. It sails 10 km due east, on 090°, to Q, and now the lighthouse is on a bearing of 300°. How far is the ship from the lighthouse?
Turn the bearings into angles inside the triangle PQL. At P, the leg to Q is on 090° and the lighthouse on 050°, so the angle QPL is 90° − 50° = 40°. At Q, the way back to P is on 090° + 180° = 270° and the lighthouse is on 300°, so the angle PQL is 300° − 270° = 30°. The angle at L is 180° − 40° − 30° = 110°.
Now a side, PQ = 10 km, is known with the angle facing it, 110°. That is the case for the sine rule. The side QL faces the 40° angle, so QL , and QL = 10 sin 40° ÷ sin 110° = 6.84 km, to 2 decimal places. In the same way, PL = 10 sin 30° ÷ sin 110° = 5.32 km.
The ship’s leg from P to Q, 10 km due east, and the line from Q to the lighthouse L on 300°, drawn to scale with a north line at each point.
The usual mistakes
Using the change of heading as the angle inside the triangle. The turn at B is 120°; the angle inside is 180° − 120° = 60°.
Adding the two legs. 8 + 5 = 13 km is the distance sailed, not the distance back to A.
Measuring a bearing from the wrong north line. A bearing from C is measured clockwise from the north line at C, so the back bearing of each leg is needed first.
Giving an angle inside the triangle as the answer for a bearing. 81.8° is the angle at C; the bearing is measured from north, and here it is 258°.
Worked example: A Ship's Two Legs and the Course Straight Back to Port
Question A ship leaves port P and sails on a bearing of 060° for 8 km to a buoy A. It then sails due south for 3 km to a point B. Take cos−1(−17) = 98.2°. (a) How far is the ship from the port? (b) On what bearing, to the nearest degree, must it sail to return straight to the port?
1.Draw a north line at A. The ship arrived on 060°, so the bearing of P from A is 060 + 180 = 240°. It leaves A due south, on 180°. The angle PAB between the two legs is 240 − 180 = 60°.
The bearing of P from A is 060 + 180 = 240° and the ship leaves on 180°, so the angle at A is 60°. 2.Two sides and the angle between them are known, so the cosine rule gives the third side: PB2 = 82 + 32 − 2 × 8 × 3 × cos 60° = 64 + 9 − 24 = 49.
The cosine rule: PB2 = 82 + 32 − 2 × 8 × 3 × cos 60° = 49. 3.(a) PB = √49 = 7 km.
(a) PB = √49 = 7 km. 4.For the angle at B the side opposite it is PA = 8 km: cos B = 32 + 72 − 822 × 3 × 7 = −642 = −17, so the angle ABP is 98.2°. The cosine is negative, so the angle is obtuse.
cos B = 9 + 49 − 6442 = −17, so the angle at B is 98.2°, obtuse. 5.A is due north of B, so the angle ABP is measured from the north line at B, turning toward the west. (b) The bearing of the port from B is 360 − 98.2 = 261.8°, which is 262° to the nearest degree. Check: the port lies a little south of due west of B, and 262° is between 180° and 270°.
(b) A is due north of B, so the bearing of P from B is 360 − 98.2 = 261.8, about 262°.
Answer: (a) 7 km; (b) 262°
Common mistakes
- Finding the angle at B with the sine rule: sin B = 8 sin 60°7 = 0.990 gives 81.8°. The sine rule cannot tell 81.8° from 98.2°, and B is obtuse here; the negative cosine is what shows it.
- Giving the bearing as 098°, the angle inside the triangle. A bearing is measured clockwise from north, and the port lies to the west of B, so the bearing is 360° minus that angle.