A two-stage journey is one non-right triangle in disguise: the start, the turning point and the finish. Find the angle inside the triangle at the turning point, use the cosine rule for the direct distance, then use an angle to reach the final bearing.
Before this, be secure on bearings and north lines and on choosing the sine or cosine rule.
What is the plan?
- Sketch three points and draw a north line at each point where a bearing starts.
- Mark the two leg lengths on the sketch.
- Find the interior angle at the turning point, using parallel north lines.
- Choose the rule: two sides and the included angle means the cosine rule.
- Find the angle at the start point with the sine rule, then add or subtract it from the first bearing.
- Check the answer against your sketch and with a rough scale drawing.
Worked example
A boat leaves port P and sails 10 km on a bearing of 040° to Q. It then sails 14 km on a bearing of 120° to R. Find the distance PR and the bearing of R from P.
Step 1, interior angle at Q: the bearing of P from Q is 040° + 180° = 220°. The bearing of R from Q is 120°. The angle between QP and QR is 220° − 120° = 100°.
Step 2, distance: two sides (10 and 14) and the included angle (100°), so use the cosine rule.
PR² = 10² + 14² − 2 × 10 × 14 × cos 100° = 296 − 280 × (−0.1736) = 296 + 48.62 = 344.6
PR = 18.56, so PR = 18.6 km.
Step 3, angle at P: sine rule, sin P / 14 = sin 100° / 18.56.
sin P = 14 × 0.9848 ÷ 18.56 = 0.7427, so angle QPR = 47.96°, which is 48.0°. It is acute, because the side opposite an obtuse angle is the longest side and the other two angles must be smaller.
Step 4, final bearing: the first leg was 040°, and R lies a further 48.0° clockwise. 40° + 48.0° = 88.0°.
Answer: PR = 18.6 km, and the bearing of R from P is 088°.
Check with components: east = 10 sin 40° + 14 sin 120° = 6.43 + 12.12 = 18.55 km. North = 10 cos 40° + 14 cos 120° = 7.66 − 7 = 0.66 km. That puts R almost due east of P, which matches 088°.
The mistake to watch for
The usual slip is using the wrong interior angle at the turning point.
Mistaken working: “The bearings are 040° and 120°, so the angle at Q is 120° − 40° = 80°.”
The student subtracted the bearings of the legs without turning the first one round. The bearing of P from Q is the back bearing 220°, not 040°.
Another common version is 40° + 120° = 160°. The correction is to draw north at Q, and measure clockwise to QP and to QR. The gap between those two arms, 220° − 120°, is the interior angle.
Check yourself
1. A hiker walks 6 km on a bearing of 000°, then 8 km on a bearing of 090°. Find the distance from the start and the final bearing from the start.
Show answer
The legs are at right angles. Distance = √(6² + 8²) = √100 = 10 km. The angle east of north is tan⁻¹(8/6) = 53.13°, so the bearing is 053°.
2. The bearing of Q from P is 050°, and the bearing of R from Q is 140°. Find angle PQR.
Show answer
The bearing of P from Q is 050° + 180° = 230°. So angle PQR = 230° − 140° = 90°.
3. A ship sails 15 km on a bearing of 090°, then 20 km on a bearing of 200°. Find the angle at the turning point and the distance from the start.
Show answer
Bearing of the start from the turning point is 270°. Angle = 270° − 200° = 70°. Cosine rule: d² = 15² + 20² − 2 × 15 × 20 × cos 70° = 625 − 600 × 0.3420 = 625 − 205.2 = 419.8. So d = 20.49, giving 20.5 km.
Where this leads next
Use the same triangle to find a triangle’s size in finding an area from two sides and an included angle. The triangle and bearings reasoning board lets you test your interior angle before calculating, and the non-calculator working trainer helps with the arithmetic. Later, check your work on the mixed practice set.
Much of the credit in a journey question come from the diagram, not the formula. Our teachers in online one-to-one Mathematics tuition look at how you sketch before they look at how you calculate.