Educerie · IB Diploma · Mathematics: analysis and approaches
Topic 3 Geometry and trigonometry · 3.3 Applications of trigonometry
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Turn a written description into a clear, labelled diagram | SL, HL | "Draw a diagram to represent this information" (1 to 2 marks) |
| Use Pythagoras' theorem and right-angled trigonometry in a context | SL, HL | a ladder, a ramp, a kite string (2 to 4 marks) |
| Use angles of elevation and depression correctly | SL, HL | "The angle of elevation of the top of the tower from A is 28°" |
| Find a height or distance from two observations | SL, HL | "Find the height of the tower" (4 to 6 marks), Paper 2 |
| Use three-figure bearings, and back bearings | SL, HL | "Find the bearing of C from A" (2 to 4 marks) |
| Use the sine and cosine rules in a bearings or navigation problem | SL, HL | "Find the distance AC" after a two-leg journey (3 to 4 marks) |
| Find right-angled triangles in a three-dimensional situation | SL, HL | a mast seen from two points in different directions (4 to 6 marks) |
Before you start
You need all of 3.2: SOH CAH TOA, the sine rule, the cosine rule and the area rule, and when to use each. You need Pythagoras' theorem, compass directions and three-figure bearings from before the course, and the angle facts for parallel lines: alternate angles are equal, and co-interior angles add to 180°. Those two facts are what make bearings and angles of depression work. The formula booklet gives the sine and cosine rules; it gives nothing specific to this page, because this page is about setting up, not about new formulas.
1The idea in one paragraph
This subtopic has no new formula. It is the skill of reading a real situation and drawing it so that the tools of 3.2 apply. Heights and distances come from right-angled triangles whenever something vertical stands on something horizontal, and from the sine and cosine rules when two observations or two legs of a journey make a triangle with no right angle. Two conventions carry most of the marks: an angle of elevation or depression is always measured from the horizontal, and a bearing is always measured clockwise from north, written with three figures. Get the diagram right and the calculation is usually two lines; get it wrong and no amount of calculation will rescue it.
2From words to a labelled diagram
The guide lists "construction of labelled diagrams from written statements" as content in its own right, and examiners award marks for it. A good diagram does four things.
- It shows every object as a point or a line: a tower is a vertical segment, a ship is a point, the sea or the ground is a horizontal line.
- It marks every right angle that the situation implies. A tower on level ground meets the ground at 90°. North–south and east–west lines meet at 90°.
- It puts every given length and angle on the diagram, in the right place, and marks the unknown with a letter.
- It is roughly to scale for angles: an angle of 12° should look small and an angle of 120° should look obtuse. This catches mistakes before they happen.
A first example: Pythagoras in context. "A ladder 6.5 m long leans against a vertical wall. Its foot is on horizontal ground, 2.5 m from the wall. Find how far up the wall the ladder reaches, and the angle the ladder makes with the ground."
The words vertical and horizontal tell you the wall meets the ground at 90°. The ladder is the hypotenuse.
That is the whole pattern: diagram, identify the right angle, choose the ratio.
3Angles of elevation and depression
The angle of elevation is the angle you look up through, from the horizontal, to see an object above you. The angle of depression is the angle you look down through, from the horizontal, to see an object below you. Figure 1 shows both for one observer on a cliff and one boat.
Elevation and depression are both measured from the horizontal. The angle of depression from the cliff to the boat equals the angle of elevation from the boat to the cliff, because the two horizontal lines are parallel and the angles are alternate.
That equality is the trick that places the angle inside the triangle. The angle of depression sits outside the triangle, between the horizontal and the line of sight; move it down to the boat as the angle of elevation and it becomes an angle of the right-angled triangle, where SOH CAH TOA can use it.
Worked example: two boats (Paper 2). From the top of a lighthouse 45 m above sea level, a keeper sees two boats due east of the lighthouse, in line with it. The angles of depression are 20° and 12°. Find the distance between the boats.
Figure 2 shows the set-up. Each boat gives its own right-angled triangle with the lighthouse, and in each the height 45 m is opposite the angle at the boat.
The smaller angle of depression belongs to the further boat. If your further boat comes out nearer, you have swapped the angles.
4A height from two observations
A classic problem: you cannot reach the foot of a tower, so you measure the angle of elevation of its top from two points in line with it, a known distance apart. There are two good methods; learn the sine-rule one, because it generalises.
Worked example (Paper 2). From a point P, the angle of elevation of the top T of a tower is 28°. A surveyor walks 50 m directly towards the tower, to Q, and the angle of elevation is now 41°. The ground is horizontal. Find the height of the tower.
Look at triangle PQT in Figure 3. Its angle at Q is 180° − 41° = 139° (angles on a straight line), so its angle at T is 180° − 28° − 139° = 13°. Quicker: the exterior angle 41° equals the sum of the two interior opposite angles, 28° + 13°.
The other method: two right-angled triangles. Let QF = x. Then tan 41° = h/x and tan 28° = h/(x + 50). Writing x = h/tan 41° and substituting gives h = 50 ÷ (1/tan 28° − 1/tan 41°) = 68.5 m. Same answer, more algebra. On Paper 2 either is fine; the sine-rule route has fewer places to slip.
5Bearings
A bearing gives a direction as an angle measured clockwise from north, written with three figures: east is 090°, south is 180°, west is 270°, and a direction 50° clockwise of north is 050°, not 50°. Figure 4(a) shows four bearings from one point.
The phrase "the bearing of B from A" means: stand at A, face north, turn clockwise until you face B. The angle you turn through is the bearing. Draw the north line at A, the point you stand at, not at B.
Back bearings. If B is on a bearing of 060° from A, then A is on a bearing of 060° + 180° = 240° from B, as in Figure 4(b). In general the back bearing is the bearing ± 180°, whichever keeps it between 000° and 360°. This works because the north lines at A and B are parallel, so the two angles made with the line AB are co-interior and add to 180°.
Worked example: a two-leg journey (Paper 1). A boat sails 8 km from A on a bearing of 070° to B, then 5 km on a bearing of 190° to C. Find the distance AC and the bearing of C from A.
The difficulty is the angle at B. Draw a north line at B, as in Figure 5. The bearing of A from B is the back bearing 070° + 180° = 250°. The boat leaves B on 190°. Both are measured clockwise from the same north line, so the angle between them is 250° − 190° = 60°.
The numbers were chosen for Paper 1. Now the bearing, which needs the angle at A (a calculator helps here, so treat this part as Paper 2).
Why add? Looking at Figure 5, C is clockwise of B as seen from A, so the bearing of C is further round than 070°. Always check that the direction on your sketch agrees with the number: 108° is a little south of east, and C is indeed south-east of A.
Bearings and right angles. When two legs are at 90° to each other, no rule is needed beyond Pythagoras. If a ship sails 12 km on 050° then 9 km on 140°, the turn is exactly 90° (140° − 050°), so the distance back to the start is √(12² + 9²) = 15 km.
6Right-angled triangles in three dimensions
Some situations are three-dimensional: a mast seen from two points in different directions, a kite above a field. In SL examinations these use only right-angled trigonometry (3.1), and the diagram does the heavy lifting.
Worked example: from a written statement (Paper 1). "A vertical mast FT stands on horizontal ground. Point A is due south of F, and the angle of elevation of T from A is 30°. Point B is due east of F, and the angle of elevation of T from B is 45°. The distance AB is 200 m. Find the height of the mast."
Build the diagram from the words, one phrase at a time, as in Figure 6. Vertical mast on horizontal ground: FT meets the ground at 90°, so triangles AFT and BFT both have their right angle at F. Due south and due east: the directions FA and FB are at 90° to each other, so triangle AFB, lying flat on the ground, has its right angle at F too.
Three right-angled triangles, two of them standing up and one lying down. The skill was entirely in seeing them.
7Where marks are lost
Measuring elevation or depression from the vertical. Both are measured from the horizontal. An angle of depression of 20° makes an angle of 70° with the vertical, and using 70° by mistake gives a wildly wrong distance.
Putting the angle of depression inside the triangle at the top. The angle of depression is between the horizontal and the line of sight. Transfer it to the object below as an angle of elevation (alternate angles) before using it in the triangle.
Two-figure bearings. Write 050°, not 50°. On some papers the bare number loses the accuracy mark.
Measuring a bearing anticlockwise, or from the wrong point. Bearings go clockwise from north at the point you are measuring from. "The bearing of B from A" means the north line is at A.
Getting the angle inside the triangle wrong in a journey. Draw a north line at every turning point, find the back bearing, and subtract. Guessing the angle from the picture is where most bearing questions go wrong.
Using the slant distance as the horizontal distance. The distance along a line of sight is the hypotenuse, not the ground distance.
Stopping at a length when a bearing was asked for. The angle inside the triangle is not the bearing. Add it to, or subtract it from, a known bearing, and check the direction on the sketch.
Rounding the first leg of a two-step method. In the two-observation problem, carry QT at full accuracy into the height. Using 104 instead of 104.34… gives 68.2 m, not 68.5 m.
8Work it right
- Read the whole question once, then draw the diagram phrase by phrase. Vertical and horizontal mean a right angle; due north, south, east and west mean right angles too.
- Put every given value on the diagram and mark the unknown.
- For elevation or depression, mark the horizontal line and measure from it. Move a depression angle to the lower object.
- For bearings, draw a north arrow at every point where a bearing is measured, and write each bearing with three figures.
- At a turning point, find the back bearing first, then the angle inside the triangle.
- Choose the tool from 3.2: right angle, SOH CAH TOA or Pythagoras; a known side–angle pair, sine rule; two sides and the included angle, cosine rule.
- Answer the question asked: a bearing, a height, a distance, with units and three significant figures.
9Try it
Marks in brackets. Q1 is Paper 1 style, no calculator. Q2 to Q5 are Paper 2 style, with a GDC.
Q1. A yacht leaves a port P and sails 6 km on a bearing of 030° to a buoy Q. It then sails 10 km on a bearing of 150° to a point R.
(a) Draw a diagram to show this information, and show that the angle PQR is 60°. 3 marks
(b) Find the exact distance PR. 3 marks
Q2. From the top of a building 32 m high, the angle of depression of a car on the level road is 17°. The car drives directly towards the building, and the angle of depression becomes 29°. Find how far the car has travelled. 5 marks
Q3. A radio mast stands on horizontal ground. From a point A, the angle of elevation of the top of the mast is 35°. From a point B, 120 m further from the mast along the same straight line through its foot, the angle of elevation is 21°.
(a) Show that the angle at the top of the mast in the triangle formed by A, B and the top is 14°. 2 marks
(b) Find the height of the mast. 4 marks
Q4. Two lifeboats leave a station S at the same time. Lifeboat A travels 14 km on a bearing of 062°. Lifeboat B travels 9 km on a bearing of 175°.
(a) Find the angle ASB. 1 mark
(b) Find the distance between the lifeboats. 3 marks
(c) Find the bearing of B from A. 4 marks
Q5. A vertical flagpole FT stands on horizontal ground. Point A is due west of F, and the angle of elevation of T from A is 40°. Point B is due south of F, and the angle of elevation of T from B is 25°. The distance AB is 150 m.
(a) Sketch a labelled diagram showing this information. 2 marks
(b) Find the height of the flagpole. 4 marks
10In one breath
Every application starts with a labelled diagram built phrase by phrase: vertical on horizontal is a right angle, due north and due east are at right angles, and every given length and angle goes on the picture before any calculation. Angles of elevation and depression are both measured from the horizontal, and they are equal because the horizontals are parallel, so an angle of depression can be moved down into the triangle as an angle of elevation. A height from two observations comes from the sine rule in the slanting triangle, with its top angle found as the difference of the two elevations, and then one right-angled triangle. Bearings are clockwise from north in three figures; the back bearing differs by 180°, and the angle at a turning point is found by drawing north there and subtracting. Then the tools of 3.2 finish the job.
Answers
Q1. (a) At Q, the back bearing of P is 030° + 180° = 210°, and R is on a bearing of 150°, so PQR = 210° − 150° = 60°. A1 for a diagram with north lines, the two legs and their lengths, M1 for the back bearing 210°, A1 for the subtraction giving 60°. This is "show that": the 210° must be seen.
(b) PR² = 6² + 10² − 2(6)(10) cos 60° = 36 + 100 − 60 = 76, so PR = √76 = 2√19 km. M1 for the cosine rule with 60° as the included angle, A1 for 76, A1 for 2√19 (√76 is also accepted as exact).
Q2. Before: d₁ = 32 ÷ tan 17° = 104.66… m. After: d₂ = 32 ÷ tan 29° = 57.72… m. Distance travelled = d₁ − d₂ = 46.9 m (3 s.f.). M1 for placing 17° and 29° as angles of elevation at the car (or equivalent), M1 for tan = 32 ÷ distance, A1 for 104.7, A1 for 57.7, A1 for 46.9. Using tan 73° and tan 61° (angles with the vertical) scores M0 for the first mark.
Q3. (a) Let M be the top of the mast. In triangle ABM, the angle at B is 21°, and the angle at A is 180° − 35° = 145°. So the angle at M is 180° − 21° − 145° = 14°. M1 for the angle 145° at A (or for using the exterior angle 35° = 21° + angle at M), A1 for 14°.
(b) Sine rule: AM ÷ sin 21° = 120 ÷ sin 14°, so AM = 120 sin 21° ÷ sin 14° = 177.76… m. Height = AM sin 35° = 102 m (3 s.f.). M1 for the sine rule with 120 facing 14°, A1 for AM = 177.8, M1 for height = AM sin 35°, A1 for 102. The two-tangent method, h = 120 ÷ (1/tan 21° − 1/tan 35°), earns full marks too.
Q4. (a) ASB = 175° − 062° = 113°. A1.
(b) AB² = 14² + 9² − 2(14)(9) cos 113° = 375.46…, so AB = 19.4 km (3 s.f.). M1 for the cosine rule with 113°, A1 for AB² ≈ 375, A1 for 19.4.
(c) Sine rule: sin SÂB ÷ 9 = sin 113° ÷ 19.376…, so SÂB = 25.3°. The bearing of S from A is 062° + 180° = 242°. B is anticlockwise of S as seen from A (B lies to the south of S, S to the south-west of A), so the bearing of B from A is 242° − 25.3° = 217° (3 s.f.). M1 for the sine rule for the angle at A, A1 for 25.3°, M1 for the back bearing 242°, A1 for 217°. An answer of 267° (adding instead of subtracting) scores M1 A1 M1 A0; the sketch shows B is south-west of A, not west.
Q5. (a) A sketch with FT vertical, A due west and B due south of F on the ground, right angles marked at F in triangles AFT, BFT and AFB, the angles 40° at A and 25° at B, and AB = 150 m. A1 for the vertical pole with both right angles at F in the upright triangles, A1 for the right angle AFB on the ground and the given values in place.
(b) AF = h ÷ tan 40°, BF = h ÷ tan 25°. Since AFB = 90°, AF² + BF² = 150², so h²(1/tan² 40° + 1/tan² 25°) = 22 500, giving h = 61.1 m (3 s.f.). M1 for AF and BF in terms of h, M1 for Pythagoras in triangle AFB, A1 for a correct equation in h, A1 for 61.1.
Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 3.3 Applications of trigonometry. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.