A survey boat leaves a pier at Hua Hin on a bearing of 062 and runs 5 km. It turns onto 145 and runs 8 km. How far is it from the pier? Subtract the bearings and you get 8.90 km, which is 1.04 km wrong.
A bearing is measured from north at the point you are standing on. At the turn the boat is looking back down the first leg, so the first leg arrives as its back bearing, 242. The angle in the triangle is 242 − 145 = 97°, not 145 − 62 = 83°.
Two north arrows, because a bearing is always measured from the north line through the point you are at.
| Rule | What it means |
|---|---|
| Clockwise from north | Not anticlockwise, and not from the nearest axis. |
| Always three figures | A bearing of 62° is written 062. Write 62 in an exam and you have written something that is not a bearing. |
| From where you are | Every point has its own north line, and all the north lines are parallel. |
The bearing of A from B, when you know the bearing of B from A. Add 180, and if that goes over 360, take 360 off.
A back bearing is never negative and never more than 360. If yours is, you have gone one step too far or not far enough.
This is where the 97° comes from. Walk the journey with your finger. At the turn, the way back to the pier is 242, and the way on is 145. The angle between those two is the angle in the triangle. Nothing about it is a guess, and you never have to remember whether to subtract or to take the answer off 180.
Two sides, 5 and 8, with 97° between them:
Both are real numbers and neither looks ridiculous, which is exactly the problem. The check is that cos 97° is negative, so the distance must come out bigger than the 9.43 km you would get from a right angle. 8.90 is smaller, so it was never going to be right.
To give the bearing of the boat from the pier, work in components. East and north, from each leg:
And √(9.003² + 4.206²) = 9.94, which agrees with the cosine rule. Two methods, one answer, which is how you know.
Your calculator cannot work out the quadrant. tan⁻¹ gives you 65.0, an acute angle, and 65 on its own would put the boat north-east of the pier. The sketch decides what to do with it. This is the one part of the question that cannot be done on the machine.
A tower stands 45 m tall. From a point 120 m away on level ground:
tan(elevation) = 45/120, so the angle of elevation is 20.6°
The angle of depression from the top of the tower back down to that point is also 20.6°. Both are measured from the horizontal, and the two horizontals are parallel, so the angles are alternate angles between parallel lines. Equal, every time.
Depression is measured from the horizontal, not from the vertical. Taking it from the vertical gives 69.4°, which is the other angle in the same triangle and will be marked wrong. If a diagram shows the angle at the top of a vertical line, check which line it is against before you use it.
The syllabus asks you to construct labelled diagrams from written statements, and this is a marked skill rather than a courtesy. Four habits that earn marks on their own:
| Do this | Because |
|---|---|
| Draw north at every point you measure from | It is the only way the angle at a turn becomes visible. |
| Mark lengths on the sides, angles inside | A length written at a vertex gets read as the wrong side. |
| Name the corners and use the names | "Angle ABC" is unambiguous, "the angle at the turn" is not. |
| Write the bearings as arcs from north, not as numbers floating near a line | An arc shows which direction it is measured in. |
The machine will do the arithmetic and will not do the quadrant. Everything that goes wrong here goes wrong in the step the machine cannot see.
When you may use it. Applications allows a calculator in every paper. Analysis does not allow one in Paper 1, so if you sit AA you must be able to run the cosine rule and the components by hand as well.
The mark people lose. Writing 65 as the bearing. It is an acute angle off a north or south line, not a bearing, and 065 points north-east while the boat is south-east. The machine has no idea which, because you only gave it two lengths. Draw the sketch, see that the boat is south and east, and take 65 off 180 to get 115. The other one is writing a bearing with two figures: it is 062, not 62.
1. A boat leaves on 062, runs 5 km, then turns onto 145. What is the angle inside the triangle at the turn, in degrees?
2. It then runs 8 km on that second bearing. How far is it from the pier, in km to 2 decimal places?
3. What is the bearing of the boat from the pier? Give it as a three-figure bearing, so type the three digits.
4. A student answers question 2 with 8.90 km. What did they do?
5. A tower is 45 m tall. From 120 m away on level ground, find the angle of elevation of its top, to 1 decimal place.
1 markA labelled diagram with north drawn at every point you measure from.
1 markThe correct angle inside the triangle, which is where most of the question is won or lost.
1 markCorrect substitution into the cosine rule or into the components.
1 markThe answer, with a bearing written as three figures.
The diagram mark is given for drawing, before any arithmetic. Students who work straight from the words lose it and then usually lose the angle as well, because the angle is the thing the diagram was going to show them.
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