How to use this menu
Six original tasks for Year 11 Extended Unit 2: two for criterion A (a short check and an end-of-unit test), one investigation for B, one communication task for C, and two real-life tasks for D that also assess C. Set at least one task for each criterion across the unit; nobody needs all six. They sit after lessons 3, 4, 5 and 6 so that assessment is spread through the unit.
Every task fits inside one lesson of about 60 minutes. A scientific calculator in degree mode is needed throughout, apart from question 1 of A1 and A2, which are done without one. The markscheme numbers are all recomputed by the unit verifier, and the exemplars were worked by hand and checked a second time.
Descriptors: Year 11 Extended is assessed with the MYP Year 5 descriptors, paraphrased below each task. Check them against your school's current guide.
The menu at a glance
| Task | Criteria | When | Time | Format |
|---|---|---|---|---|
| A1: Quick check: angles, graphs and equations | A | After lesson 4 | 25 minutes | Individual, written, in class. 22 marks. Question 1 is done without a calculator. |
| B1: The total of three sines | B | Mid-unit, after lesson 3 | 50 minutes | In class. Students may share the values they calculate in pairs, but each writes up their own findings and reasons. |
| C1: Explain it to someone who missed it | C | After lesson 5 | 35 minutes | Individual, written, in class or as homework. One page of words and diagrams. |
| A2: End-of-unit test | A | End of unit, after lesson 6 | 50 minutes | Individual, written, in class, one lesson. 31 marks in three sections. Question 1 is done without a calculator. |
| D1: Tide times at an island pier | D and C | End of unit, after lesson 6 | 50 minutes | Individual, in class, one lesson. A written solution with explanations and a short note to the ferry operator. |
| D2: Two boats leave the harbour | D and C | End of unit, after lesson 6 | 50 minutes | Individual, in class, one lesson. A written solution with explanations and a short note to the harbour radio operator. |
D1 and D2 are alternatives. Both assess criteria D and C: D1 models a tide with a cosine, D2 uses the sine and cosine rules and bearings for two boats. Set one, not both.
The unit's inquiry
- Key concept. Relationships: how an angle, a length and a height depend on one another in a circle, a triangle and a wave.
- Related concepts. Models, Pattern, Space
- Global context. Globalization and sustainability: how people on coasts and at sea use repeating cycles and distances to plan safe journeys.
- Statement of inquiry. Sine and cosine link angles to heights and distances, so one rule can describe a cycle, locate a point and decide whether a journey is possible.
MYP Year 5 descriptors, paraphrased
Each task's clarification, on its student sheet below, says what these levels look like in that task.
| Level | Criterion A: Knowing and understanding |
|---|---|
| 1–2 | Selects suitable mathematics for simple problems in familiar situations and applies it with some success. |
| 3–4 | Selects suitable mathematics for more complex problems in familiar situations and applies it successfully. |
| 5–6 | Selects suitable mathematics for challenging problems in familiar situations, applies it successfully and generally solves these problems correctly in a range of contexts. |
| 7–8 | Selects suitable mathematics for challenging problems in both familiar and unfamiliar situations, applies it successfully and generally solves these problems correctly in a range of contexts. |
| Level | Criterion B: Investigating patterns |
|---|---|
| 1–2 | With support, applies problem-solving techniques to find simple patterns and states predictions that fit them. |
| 3–4 | Applies problem-solving techniques to find simple patterns and suggests general rules that fit the findings. |
| 5–6 | Applies problem-solving techniques to find complex patterns, describes them as general rules that fit the findings and verifies that the rules hold. |
| 7–8 | Selects and applies problem-solving techniques to find complex patterns, describes them as general rules that fit correct findings, and proves, or verifies and justifies, those rules. |
| Level | Criterion C: Communicating |
|---|---|
| 1–2 | Uses limited mathematical language and limited forms of representation; the reasoning is hard to follow. |
| 3–4 | Uses some appropriate mathematical language and forms of representation; the reasoning can be followed and the work is adequately organized. |
| 5–6 | Usually uses appropriate language, moves between forms of representation with some success, and the reasoning is complete and coherent in a mostly logical structure. |
| 7–8 | Consistently uses appropriate language, moves effectively between forms of representation, and the reasoning is complete, coherent and concise in a logical structure. |
| Level | Criterion D: Applying mathematics in real-life contexts |
|---|---|
| 1–2 | Identifies some relevant elements of the real-life situation and selects, with some success, strategies to model it. |
| 3–4 | Identifies the relevant elements, selects adequate strategies to model the situation, applies them to reach a solution and discusses whether it makes sense. |
| 5–6 | Identifies the relevant elements, selects adequate modelling strategies and applies them to reach a valid solution, explains the degree of accuracy and explains whether the solution makes sense. |
| 7–8 | Identifies the relevant elements, selects appropriate modelling strategies and applies them successfully to reach a correct solution, justifies the degree of accuracy and justifies whether the solution makes sense. |
Task A1: Quick check: angles, graphs and equations
Why set it. A short check of lessons 1 to 4 while there is still time to reteach: exact values and symmetry, amplitude and period, and solving equations in an interval. Questions 7 and 8 are less routine, so the top band is open to everyone.
When: After lesson 4. Time: 25 minutes. Format: Individual, written, in class. 22 marks. Question 1 is done without a calculator.
The student sheet, with its task-specific clarification
When: After lesson 4. Time: 25 minutes. Calculator: scientific calculator. Criterion: A, knowing and understanding.
Show your working. Angles are in degrees. Question 1 is done without a calculator; use one for the rest and give angles to 1 decimal place unless a question says otherwise.
Exact values and symmetry
- Without a calculator, write down the exact value of (a) cos 300° (b) tan 135°. (2 marks)
- Given that sin 25° = 0.4226, write down the values of sin 155°, sin 205° and sin 335°. (3 marks)
Graphs
- For the graph of y = 3 cos 4x + 1, state the amplitude, the period and the least value of y. (3 marks)
- The graph of y = a sin bx, with a and b positive, has a greatest value of 5. It passes through the origin and next crosses the x-axis at x = 60°. Find a and b. (3 marks)
Equations in an interval
- Solve sin x = 0.35 for 0° ≤ x ≤ 360°. (3 marks)
- Solve 4 cos x + 1 = −1 for 0° ≤ x ≤ 360°. (3 marks)
- Solve 2 sin x = cos x for 0° ≤ x ≤ 360°. (3 marks)
- How many solutions does cos x = 0.8 have for 0° ≤ x ≤ 720°? Explain how you know without solving it. (2 marks)
How your work is judged
| Level | Criterion A in task A1: what it looks like |
|---|---|
| 1–2 | You answer some of the routine questions correctly, for example an exact value or the amplitude of a graph. |
| 3–4 | You answer most of the routine questions correctly, including a solution of an equation, and give at least one correct solution to the others. |
| 5–6 | You answer the routine questions correctly with clear methods, and find every solution to questions 5 to 7 in the interval. |
| 7–8 | You answer questions 1 to 7 correctly and explain your reasoning in question 8, including why each extra turn of the graph adds solutions. |
Markscheme
| Task A1 question | Answer and marking |
|---|---|
| 1 | (a) 1 ÷ 2 (b) −1. 1 mark each. cos 300° = cos 60° (fourth quadrant, cosine positive); tan 135° = −tan 45° (second quadrant, tangent negative). Sign errors are the common loss: −1 ÷ 2 for (a), or +1 for (b). |
| 2 | 0.4226, −0.4226, −0.4226. 1 mark each. 155° = 180° − 25°, 205° = 180° + 25°, 335° = 360° − 25°. The sine is positive only in the first two quadrants. |
| 3 | Amplitude 3, period 90°, least value −2. 1 mark each. The period is 360 ÷ 4; the least value is 1 − 3. A period of 360° or a least value of −3 ignores the 4 or the 1. |
| 4 | a = 5, b = 3. 1 mark for a, 2 for b: the graph crosses the axis half a period in, so the period is 120° and b = 360 ÷ 120 = 3. b = 6 from using 60° as the period earns 1 mark for the idea. |
| 5 | 20.5° and 159.5°. 1 mark for the calculator angle 20.5°, 1 for the second angle 180° − 20.5°, 1 for both given to 1 decimal place. 200.5° adds 180°, which is the rule for the tangent. |
| 6 | 120° and 240°. 1 mark for cos x = −1 ÷ 2, 1 each for the solutions: the calculator gives 120° and the second is 360° − 120°. |
| 7 | 26.6° and 206.6°. 1 mark for dividing by cos x to reach tan x = 0.5, 1 for 26.6°, 1 for 206.6° (26.6° + 180°). 153.4° comes from using the sine’s second solution (180° − x) for the tangent. Dividing the wrong way gives tan x = 2 (63.4° and 243.4°): that earns the method mark for dividing by cos x only. |
| 8 | 4 solutions. 1 mark for the answer. 1 mark for the reason: in each turn of 360° the line y = 0.8 meets the cosine curve twice (once going down, once coming back up), and 720° is two turns. The solutions are 36.9°, 323.1°, 396.9° and 683.1°. |
| Marks to a level | Total 22. 1 to 6 marks: 1 to 2. 7 to 12: 3 to 4. 13 to 18: 5 to 6. 19 to 22, with question 7 fully correct and a reason given in question 8: 7 to 8. Use the marks as a guide, then check the profile against the clarification. |
Exemplar answers
| Task A1 response | What it looks like |
|---|---|
| Level 3 to 4 profile (8 of 22 marks) | Q1 (a) 1 ÷ 2 and (b) 1 (1 mark: the sign of tan 135° is wrong). Q2 0.4226 for all three (1 mark). Q3 amplitude 3, period 90°, least value 4 (2 marks). Q4 a = 5 only (1 mark). Q5 20.5° (1 mark: no second angle). Q6 120° (2 marks). Q7 no marks. Q8 “2” with no reason (0). Correct routine skills in places, but signs and second solutions are missing. |
| Level 5 to 6 profile (19 of 22 marks) | Q1 and Q2 all correct (5). Q3 correct (3). Q4 a = 5, b = 3 (3). Q5 20.5° and 159.5° (3). Q6 120° and 240° (3). Q7 “tan x = 2”: the division was done the wrong way, 63.4° and 243.4° (1 mark for the method of dividing by cos x). Q8 “4, the graph has two turns” (1). Sound on every routine item; the equation that needs rearranging and the explanation hold the marks back. 19 marks reach the top mark band, but question 7 is not fully correct, so the level is 5 to 6. |
Differentiation
A way in for anyone stuck
- Allow a copy of the unit circle diagram with the four quadrants and the signs written in, as in lesson 1.
- Give a pencilled sketch of the sine graph from 0° to 360° for question 5 so the second solution can be read from the symmetry.
- For question 7, prompt with “What does sin x ÷ cos x equal?” rather than showing the step.
Stretch
- Solve sin x = 0.35 for 0° ≤ x ≤ 720°. (20.5°, 159.5°, 380.5°, 519.5°)
- How many solutions does sin 2x = 0.35 have for 0° ≤ x ≤ 360°, and why? (4: the graph has two waves in 360°.)
Students learning in English as an additional language
| Word in task A1 | Plain meaning |
|---|---|
| amplitude | half the distance between the highest and the lowest value of a wave |
| period | the angle after which the graph starts to repeat |
| solution | a value of x that makes the equation true |
| interval | all the values between two limits, here 0° and 360° |
| quadrant | one quarter of the circle: first, second, third or fourth |
| exact value | an answer written with fractions and roots, not rounded decimals |
Sentence starters for task A1:
- The angle is in the … quadrant, so the sign is …
- The second solution is … because the graph is symmetrical about …
- The period is 360 ÷ … = …
Task B1: The total of three sines
Why set it. A real pattern hunt built on lessons 1 to 3. The sheet never names the pattern: students organize values, find it, state it as a general rule, test and extend it, and then justify it for every angle, which is where the evidence for the top band comes from.
When: Mid-unit, after lesson 3. Time: 50 minutes. Format: In class. Students may share the values they calculate in pairs, but each writes up their own findings and reasons.
The student sheet, with its task-specific clarification
In MYP Year 5 a prompt such as “try more values of x” is acceptable at 1 to 2 and 3 to 4; note it on the student’s work. The justification in questions 8 and 9 is the evidence for 7 to 8.
When: Mid-unit, after lesson 3. Time: 50 minutes. Calculator: scientific calculator. Criterion: B, investigating patterns.
Choose any angle x and look at the three angles x, x + 120° and x + 240°. Call the total of their sines T(x), so that
T(x) = sin x + sin (x + 120°) + sin (x + 240°).
This task is about what T(x) does as x changes, how far you can trust what you see, and why. Use a calculator in degree mode and give values to 3 decimal places.
Part 1. Collect values
- Work out T(x) for x = 0°, 20°, 50° and 75°. Record each of the three sines and the total in a table you design yourself.
- Choose two more values of x of your own, one of them larger than 90°, and add them to your table.
Part 2. Describe and predict
- Describe what you notice about T(x), in words and then as a general statement. Say what evidence you have.
- Use your statement to predict T(200°) and T(310°). Then work out both to check your predictions.
Part 3. Test and extend
- Does your statement still hold if the sines are replaced by cosines? Test it for at least three values of x.
- Investigate the total of the sines of (a) four angles x, x + 90°, x + 180° and x + 270° (b) three angles x, x + 100° and x + 200°. In each case say what changes and what stays the same.
- Choose a number of angles and a spacing of your own. Predict what the total of their sines will do, then test your prediction. State the conditions under which your general statement holds.
Part 4. Justify
- For the four angles in question 6(a), use what you know about sin (x + 180°) from lesson 1 to explain why your result holds for every x, not only the ones you tried.
- Explain, using the unit circle, why your result for x, x + 120° and x + 240° holds for every x. Say what a table of values can show and what it cannot.
How your work is judged
| Level | Criterion B in task B1: what it looks like |
|---|---|
| 1–2 | With help, you work out some values of T(x), notice something about them and make a prediction for T(200°). |
| 3–4 | You complete a table for question 1, suggest a general statement that fits your values and use it to predict T(200°) and T(310°). |
| 5–6 | You organize your values clearly, state a general rule, test whether it also holds for the cosines, and verify it with new values and with the changed angles in question 6. |
| 7–8 | You state a general rule that fits all your findings, say when it holds and when it does not, test it on a case you chose, and justify it with a reason that works for every x, not only the values you tried. |
Markscheme
| Task B1 question | Answer and marking |
|---|---|
| The pattern (do not share it before the task) | For any x, sin x + sin (x + 120°) + sin (x + 240°) = 0, and the same for the cosines. Any number of angles spread evenly round the full circle gives a total of 0. It fails when the angles do not close the circle: three angles 100° apart give a total that changes with x. |
| 1 | T(0°) = 0 + 0.866 − 0.866 = 0.000; T(20°) = 0.342 + 0.643 − 0.985 = 0.000; T(50°) = 0.766 + 0.174 − 0.940 = 0.000; T(75°) = 0.966 − 0.259 − 0.707 = 0.000. Credit a table that shows each term. Rounding can leave a total of 0.001 or −0.001: a student who writes “about 0” and says why is showing good judgement. |
| 2 | Any extra x with the three sines and a total of 0.000. For example T(130°) = 0.766 − 0.940 + 0.174 = 0.000. A value over 90° should show at least one negative term. |
| 3 | Any correct description of the total being 0 for every x tried. A statement about all x with the evidence named is evidence for 5 to 6; “it is 0 every time” with no further words is evidence for 3 to 4. |
| 4 | T(200°) = 0 and T(310°) = 0. The terms are −0.342 − 0.643 + 0.985 for 200° (sin 200°, sin 320°, sin 440°) and −0.766 + 0.940 − 0.174 for 310° (sin 310°, sin 430°, sin 550°). |
| 5 | The cosines also add up to 0. For example at x = 0°: 1 − 0.5 − 0.5 = 0. Credit three tested values and the conclusion that the same rule holds. |
| 6 | (a) Total 0 for every x tried. For example x = 20°: 0.342 + 0.940 − 0.342 − 0.940 = 0. (b) Total 0.643 at x = 0°, 0.420 at x = 40°, −0.113 at x = 90°: it changes with x. What stays the same: four angles 90° apart still give 0. What changes: three angles 100° apart do not close the circle, so the total varies. |
| 7 | Credit any well-chosen case with a prediction made first and then tested, for example five angles 72° apart (total 0), six angles 60° apart (0) or two angles 180° apart (0). A statement of the condition, that the angles are spread evenly round the full 360°, is evidence for 7 to 8. A student who chooses three angles 90° apart finds a total that varies (1.000 at x = 0°, 0.707 at x = 45°) and says so: also full credit. |
| 8 | sin (x + 180°) = −sin x and sin (x + 270°) = −sin (x + 90°), so the first and third terms are negatives of each other, and so are the second and fourth: the total is 0 for every x. A table alone cannot show this for every x. |
| 9 | The three angles give three points equally spaced round the unit circle, which are the corners of an equilateral triangle centred on the origin. The sine of each angle is the height of its point. In an equilateral triangle the centre of the circle through the corners is also the average of the three corners, so the average height, and with it the total, is 0. (Equivalently, turning the picture through 120° about the origin leaves it unchanged, so the total is unchanged by the turn, and the only fixed point is the origin.) A table can show that the rule works for the values chosen but cannot cover every x. Accept any reasoning that works for all x. |
Exemplar answers
| Task B1 response | What it looks like |
|---|---|
| Level 3 to 4 | “I worked out the table and the total is 0 each time. I think it is always 0. T(200°) will be 0.” Both predictions are correct and the table is complete, but the statement is made with no comment on rounding, cosines or a different set of angles, and nothing is justified. |
| Level 7 to 8 | “For every x I tried, the three sines add up to 0 to 3 decimal places, and the three cosines do too. I tested four angles 90° apart and five angles 72° apart: also 0. With three angles 100° apart the total was 0.643, 0.420 and −0.113 for x = 0°, 40° and 90°, so it only works when the angles are spread evenly round the whole circle. A table cannot prove it for every x. The three angles are the corners of an equilateral triangle in the unit circle centred on the origin; the sines are the heights of the corners; the heights of a triangle centred on the origin add up to 0.” |
Differentiation
A way in for anyone stuck
- Give a table outline with the headings x, sin x, sin (x + 120°), sin (x + 240°) and total, and let the student fill it in.
- Allow a copy of the unit circle diagram so the heights of the three points can be read before they are calculated.
- Prompt with a question, never the rule: “What do you notice about the last column?”
Stretch
- Work out sin2 x + sin2 (x + 120°) + sin2 (x + 240°) for several values of x. What do you find? (1.5 every time.)
- Find a spacing s, other than 120° and 240°, for which x, x + s and x + 2s have a total of sines that is 0 for every x. (None: only s = 120° or 240°, up to whole turns.)
Students learning in English as an additional language
| Word in task B1 | Plain meaning |
|---|---|
| total | the answer when numbers are added |
| predict | say what will happen before you check |
| general statement | a sentence that is true for every value, not only the ones you tried |
| evidence | the results that show your statement is true |
| justify | give a reason that works for every case |
| spacing | the gap between one angle and the next |
Sentence starters for task B1:
- For every value of x that I tried, the total …
- My statement fits the values because …
- A table of values cannot show … because …
Task C1: Explain it to someone who missed it
Why set it. Criterion C on its own: notation, diagrams and a line of reasoning a classmate can follow, built around two real errors from lessons 4 and 5.
When: After lesson 5. Time: 35 minutes. Format: Individual, written, in class or as homework. One page of words and diagrams.
The student sheet, with its task-specific clarification
When: After lesson 5. Time: 35 minutes. Calculator: scientific calculator. Criterion: C, communicating.
Mei was away for lessons 4 and 5. She copied this working from a classmate, Tomas, and it has mistakes in it.
Tomas’s working
(a) Solve sin x = 0.3 for 0° ≤ x ≤ 360°.
sin x = 0.3 = 17.5°. The sine repeats every 180°, so the other solution is 17.5° + 180° = 197.5°.
(b) In triangle PQR, p = 8 cm, q = 10 cm and angle P = 40°. Find angle Q.
sin Q ÷ 10 = sin 40° ÷ 8, so sin Q = 0.803 and Q = 53.5°. There is one triangle.
Write a one-page explanation for Mei. She should be able to follow it on her own, without asking anyone.
Part 1. Put it right
- Find every mistake in Tomas’s working. For each one, say what he did, why it is wrong and write the correct working.
Part 2. Teach the idea
- Explain to Mei how to find all the solutions of an equation of the form sin x = k between 0° and 360°. Use a sketch of the sine graph or a unit circle diagram, for k = 0.3 and for k = −0.6.
- Explain to Mei how the sine rule can give two possible triangles when two sides and an angle that is not between them are known. Use Tomas’s triangle in a labelled diagram.
Part 3. Check your explanation
- Make up one triangle where the sine rule gives a second angle that is not possible, so that only one triangle fits. Say how a reader can tell.
- Read your page as if you were Mei. Check that you used the degree sign, an equals sign only between things that are equal, a labelled diagram, and the words solution, symmetry, sine rule and ambiguous.
How your work is judged
| Level | Criterion C in task C1: what it looks like |
|---|---|
| 1–2 | Some of your explanation can be followed, but key steps or words are missing and the diagrams or notation are confusing. |
| 3–4 | You explain both parts using some correct words and notation, in a sensible order, so Mei could follow most of it. |
| 5–6 | Your steps are complete and ordered, with a labelled sketch or diagram, degree signs and mostly correct notation; Mei could follow it without help. |
| 7–8 | Every step is explained concisely, your diagrams and working match your words, notation is always correct, and the page is ordered so that Mei could learn from it alone. |
Markscheme
| Task C1 question | Answer and marking |
|---|---|
| The mistake in (a) | Two errors. First, “sin x = 0.3 = 17.5°” has an equals sign between a ratio and an angle. Second, the other solution is 180° − 17.5° = 162.5°, by symmetry about 90°; adding 180° is the rule for the tangent, whose graph repeats every 180°. Both solutions are 17.5° and 162.5°. |
| The mistake in (b) | Tomas stopped at the calculator angle. sin Q = 10 sin 40° ÷ 8 = 0.803, so Q = 53.5° or Q = 180° − 53.5° = 126.5°. Both are possible because 40° + 126.5° = 166.5° is less than 180°, so there are two triangles, with angle R = 86.5° or 13.5°. |
| The sketch for k = −0.6 (Part 2) | The calculator gives −36.9°. The solutions in the interval are 180° + 36.9° = 216.9° and 360° − 36.9° = 323.1°, both in the third and fourth quadrants where the sine is negative. |
| Own triangle (Part 3) | A triangle where adding the second angle to the given angle makes more than 180°. Example: a = 10, b = 8, A = 50°: sin B = 8 sin 50° ÷ 10 = 0.613, B = 37.8° or 142.2°; but 50° + 142.2° = 192.2° is more than 180°, so only one triangle fits. (The side opposite the given angle is the longer side.) |
| How to level | Level the page as a whole against the clarification, not by counting corrections. Notation, labelled diagrams and ordering decide between 5 to 6 and 7 to 8 more often than the corrections do. A slip in arithmetic counts only where it breaks the line of reasoning Mei has to follow; accuracy itself is criterion A. |
Exemplar answers
| Task C1 response | What it looks like |
|---|---|
| Level 3 to 4 extract | “Tomas is wrong. The other angle is 180 − 17.5 = 162.5. In b there are two answers 53.5 and 126.5 because the sine is also positive in the second quadrant.” The corrections are right, but there are no degree signs, no diagram and no steps showing where 126.5 comes from; the reader has to supply the reasoning. |
| Level 7 to 8 extract | “Step 1, sketch y = sin x and the line y = 0.3. They cross at the calculator angle 17.5° and again at 180° − 17.5° = 162.5°, because the curve is symmetrical about x = 90°. Tomas added 180°, which is the rule for tan x, whose graph repeats every 180°. Step 2, in triangle PQR, sin Q = 0.803 gives Q = 53.5° or Q = 126.5°; the second is allowed because 40° + 126.5° < 180°.” Each step is on its own line, with a labelled sketch, and no equals sign joins unequal things. |
Differentiation
A way in for anyone stuck
- Give the graph sketch and the unit circle diagram from lesson 4 as a reference card.
- Let students explain aloud to a partner first, then write down what they said.
- Offer the frame: “Tomas wrote …. This is wrong because …. The correct working is ….”
Stretch
- Explain why the same two angles, 17.5° and 162.5°, are both solutions of sin x = 0.3 but a different pair of angles is needed in each triangle in Tomas’s part (b).
- Write a second version of the page for a student two years younger. What did you change, and why?
Students learning in English as an additional language
| Word in task C1 | Plain meaning |
|---|---|
| symmetry | the graph looks the same on both sides of a line, so one solution gives another |
| ambiguous | possible in two different ways |
| mistake | a step that is not correct |
| explain | say what you did and why |
| labelled diagram | a picture with the names and values written on it |
Sentence starters for task C1:
- Tomas wrote …, but …
- The second solution is … because the graph …
- I know there are two triangles because …
Task A2: End-of-unit test
Why set it. The whole unit at three levels of demand. Section 3 is unfamiliar on purpose: it is where the evidence for 7 to 8 comes from.
When: End of unit, after lesson 6. Time: 50 minutes. Format: Individual, written, in class, one lesson. 31 marks in three sections. Question 1 is done without a calculator.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 6. Time: 50 minutes. Calculator: scientific calculator. Criterion: A, knowing and understanding.
Three sections, from familiar questions to ones you have not met in this form. Show your working everywhere: it earns credit even when an answer slips. Angles are in degrees; give lengths to 3 significant figures and angles to 1 decimal place unless a question says otherwise. Question 1 is done without a calculator.
Section 1. Familiar questions
- Write down the exact value of (a) sin 240° (b) cos 135°. (2 marks)
- For the graph of y = 2 sin 3x, state the amplitude and the period, and say how many complete waves there are from 0° to 360°. (3 marks)
- Solve 5 sin x = 2 for 0° ≤ x ≤ 360°. (3 marks)
- In triangle ABC, A = 48°, B = 71° and a = 9.2 cm. Find the length b. (3 marks)
Section 2. More complex questions
- A wheel at a night market has a height above the ground, in metres, of h = 9 − 8 cos 30t, where t is the time in minutes after the wheel starts. (a) Write down the least and the greatest height. (b) Write down the time for one complete turn. (c) Find the first time the height is 12 m, to the nearest tenth of a minute. (6 marks)
- A triangle has sides 12 cm, 15 cm and 19 cm. Find its largest angle, and then its area. (4 marks)
Section 3. Unfamiliar questions
- In triangle XYZ, XY = 9 cm, YZ = 6 cm and angle X = 35°. Show that two different triangles fit this information, and find both possible values of angle Z. (4 marks)
- Given that cos 50° = 0.6428, find all the values of x between 0° and 360° for which sin x = 0.6428. Explain your reasoning. (3 marks)
- Two sides of a triangle are 10 cm and 14 cm and its area is 35 cm2. Find the two possible sizes of the angle between the sides. (3 marks)
How your work is judged
| Level | Criterion A in task A2: what it looks like |
|---|---|
| 1–2 | You answer some Section 1 questions correctly, for example an exact value or the amplitude of a graph. |
| 3–4 | You answer most of Section 1 correctly and make progress in Section 2. |
| 5–6 | You answer Section 1 and most of Section 2 correctly, with clear methods, and make a start in Section 3. |
| 7–8 | You answer Sections 1 and 2 correctly and solve most of Section 3, explaining your reasoning. |
Markscheme
| Task A2 question | Answer and marking |
|---|---|
| 1 | (a) −√3 ÷ 2 (b) −√2 ÷ 2. 1 mark each. 240° = 180° + 60° (third quadrant, sine negative); 135° = 180° − 45° (second quadrant, cosine negative). Positive values lose the sign mark. |
| 2 | Amplitude 2, period 120°, 3 waves. 1 mark each. 360 ÷ 3 = 120, and 360 ÷ 120 = 3. 360° for the period is the unchanged sine. |
| 3 | 23.6° and 156.4°. 1 mark for sin x = 0.4, 1 for 23.6°, 1 for 180° − 23.6° = 156.4°. 203.6° adds 180°, which is the rule for the tangent. |
| 4 | b = 11.7 cm. 1 mark for the sine rule set-up b ÷ sin 71° = 9.2 ÷ sin 48°, 1 for the rearrangement, 1 for the answer. 7.2 comes from putting the sines the wrong way up. |
| 5 | (a) 1 m and 17 m (b) 12 minutes (c) 3.7 minutes. (a) 9 − 8 and 9 + 8, 1 mark each. (b) 360 ÷ 30, 1 mark. (c) 9 − 8 cos 30t = 12 gives cos 30t = −0.375 (1 mark), 30t = 112.0° (1 mark), t = 3.7 (1 mark). 8.3 minutes (30t = 248.0°) is the second time the height is 12 m, not the first. |
| 6 | Largest angle 88.7°; area 90.0 cm2. The largest angle is opposite 19 cm. 1 mark for cos C = (122 + 152 − 192) ÷ (2 × 12 × 15), 1 for cos C = 8 ÷ 360, 1 for 88.7°, 1 for the area (1 ÷ 2) × 12 × 15 × sin 88.7° = 90.0 (89.98). 52.1° is the angle opposite 15 cm: the wrong angle found. |
| 7 | Z = 59.4° or Z = 120.6°. sin Z = 9 sin 35° ÷ 6 = 0.860 (1 mark), Z = 59.4° (1), Z = 180° − 59.4° = 120.6° (1). The reason (1 mark): 35° + 120.6° = 155.6° is less than 180°, so both triangles exist, with angle Y = 85.6° or 24.4°. |
| 8 | x = 40° and x = 140°. 1 mark for recognising cos 50° = sin 40° (the cosine is the sine of the complementary angle), 1 for 40°, 1 for 140° = 180° − 40°. A student who finds 40° with the calculator and then uses the symmetry, with no link from cos 50° to sin 40°, earns the marks for 40° and 140° but not the reasoning mark. 50° and 130° come from confusing sine and cosine. |
| 9 | 30° or 150°. 1 mark for area = (1 ÷ 2) × 10 × 14 × sin C, 1 for sin C = 0.5, 1 for both angles (180° − 30° = 150°). 30° alone loses the last mark. |
| Marks to a level | Total 31. 1 to 7 marks: 1 to 2. 8 to 16: 3 to 4. 17 to 24: 5 to 6. 25 to 31, with at least 7 of Section 3’s 10 marks: 7 to 8. Check the profile against the clarification before settling the level. |
Exemplar answers
| Task A2 response | What it looks like |
|---|---|
| Level 3 to 4 profile (16 of 31 marks) | Q1 (a) −√3 ÷ 2, (b) 1 ÷ √2 (1 mark: sign missing). Q2 amplitude 2, period 120°, 6 waves (2). Q3 23.6° only (2). Q4 b = 11.7 (3). Q5 (a) 1 and 17, (b) 12 (3), (c) not attempted. Q6 88.7° found, no area (3). Q7 Z = 59.4° only (2). Q8 and Q9 not attempted. Familiar items mostly sound; second solutions and the unfamiliar section are missing. |
| Level 5 to 6 profile (27 of 31 marks) | Section 1 all correct (11). Section 2 Q5 all three parts correct, with 30t = 112.0° (6); Q6 88.7° and 89.98 cm2 (4). Section 3 Q7 59.4° and 120.6° with no check that both triangles exist (3). Q8 40° and 140° found by calculator and symmetry, with no reasoning about cos 50° (2). Q9 ½ × 10 × 14 × sin C = 35 written correctly (1), then sin C = 35 ÷ 140 = 0.25, dropping the half, and C = 14.5° (no further marks). 27 marks, but only 6 of Section 3’s 10, so 5 to 6 rather than 7 to 8. |
Differentiation
A way in for anyone stuck
- Allow a formula sheet with the sine rule, the cosine rule and the area formula, as in lessons 5 and 6.
- Give a copy of the unit circle diagram from lesson 1 for question 1.
- For question 5, prompt with “What is the value of cos 30t when the height is 12?”
Stretch
- For question 7, find the least and the greatest value of YZ for which there are two triangles with XY = 9 cm and X = 35°. (Greater than 9 sin 35° = 5.16 cm and less than 9 cm.)
- In question 5, for how many minutes in each turn is the height above 12 m? (4.5 minutes.)
Students learning in English as an additional language
| Word in task A2 | Plain meaning |
|---|---|
| included angle | the angle between two given sides |
| height above the ground | how far up from the ground something is |
| significant figures | the digits that matter in a number, counted from the first non-zero digit |
Sentence starters for task A2:
- The sine rule gives … = …, so the angle is … or …
- The largest angle is opposite the … side, so …
- Both triangles exist because … is less than 180°.
Task D1: Tide times at an island pier
Why set it. A periodic model used to make a timetable decision, with a test of how far the model can be trusted. An alternative to D2: set one of them.
When: End of unit, after lesson 6. Time: 50 minutes. Format: Individual, in class, one lesson. A written solution with explanations and a short note to the ferry operator.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 6. Time: 50 minutes. Calculator: scientific calculator. Criteria: D, applying mathematics in real-life contexts; C, communicating.
A ferry operator runs small boats between the mainland and a pier on an island in the Andaman Sea. At the end of the pier the depth of water, d metres, at t hours after high tide is modelled by
d = 2.6 + 1.2 cos 30t, where 30t is an angle in degrees and t is between 0 and 12.
A boat can dock only when the depth is at least 2.3 m. High tide on the day of this task is at 05:30.
Part 1. Understand the model
- State the greatest and the least depth the model gives, and the time between one high tide and the next. Say what each number in the model tells you about the tide.
- Find the depth of water 2 hours after high tide.
Part 2. Docking
- Find the times, to the nearest minute after high tide, between which boats cannot dock in the 12 hours after the first high tide. Write them as clock times on the day.
- Boats are due to arrive at 07:00, 08:30, 10:00, 11:30, 13:00, 14:30 and 16:00. Which of these can dock? Justify your answer.
Part 3. Does it make sense?
- The model uses exactly 12 hours between high tides. A real tide cycle is about 12 hours 25 minutes. How would this change the times in question 3, and the next unsafe window later in the day, and how far can you trust the timetable?
- Write a short note to the ferry operator with your recommendations, so that someone who has not seen your working can understand them.
How your work is judged
| Level | Criterion D in task D1: what it looks like |
|---|---|
| 1–2 | You find some of the right numbers, for example the greatest and least depth or the depth at one time, but there is no decision about docking. |
| 3–4 | You find the depths, the time when docking is impossible and which arrivals can dock, and say whether your answers make sense. |
| 5–6 | You choose suitable methods for every part, reach correct answers, explain how accurate the times are and whether the timetable is sensible, and describe how the 12 hour model differs from a real tide. |
| 7–8 | You choose and use suitable methods throughout, reach correct answers, explain how accurate each time is and why, and justify whether your recommendation is reasonable for a real tide. |
| Level | Criterion C in task D1: what it looks like |
|---|---|
| 1–2 | Some working is shown, but it is hard to see how you reached the times or which arrivals can dock. |
| 3–4 | Your working for the times can be followed, with some correct notation, units and clock times. |
| 5–6 | Your working is clear and in order, usually with correct units, degree signs and clock times, and your note to the operator is mostly well organized. |
| 7–8 | Your working is clear, concise and well ordered, notation and units are always correct, and your note to the operator makes sense on its own. |
Markscheme
| Task D1 question | Answer and marking |
|---|---|
| 1 | Greatest 3.8 m, least 1.4 m, 12 hours. 2.6 is the mean depth, 1.2 m is the amplitude (the depth varies 1.2 m either side of the mean), and 360 ÷ 30 = 12 hours is the period. Credit explanations in words. |
| 2 | 3.2 m. 2.6 + 1.2 cos 60° = 2.6 + 0.6. 3.8 m is the value at t = 0. |
| 3 | From 3 h 29 min to 8 h 31 min after high tide: 08:59 to 14:01. 2.6 + 1.2 cos 30t = 2.3 gives cos 30t = −0.25, 30t = 104.5° and 255.5° (360° − 104.5°), so t = 3.483 and 8.517 hours: 209 minutes and 511 minutes. The depth is below 2.3 m between these times (1 mark for each time, 1 for the interval, 1 for the clock times). |
| 4 | 07:00 (3.45 m), 08:30 (2.6 m), 14:30 (2.6 m) and 16:00 (3.45 m) can dock; 10:00 (1.75 m), 11:30 (1.4 m) and 13:00 (1.75 m) cannot. The times after high tide are 1.5, 3, 4.5, 6, 7.5, 9 and 10.5 hours. Credit comparison with the interval from question 3 or a calculation of each depth. |
| 5 | Every time after high tide stretches by 12 h 25 min ÷ 12 h = 1.035: the unsafe window becomes about 3 h 36 min to 8 h 49 min after high tide, 09:06 to 14:19, and the next high tide is at 17:55, not 17:30. The same arrivals can dock, but 14:30 is now only about 2.4 m, so it is borderline, and each later cycle drifts a further 25 minutes. Credit a statement that the model repeats exactly but a real tide drifts, and a recommended safety margin of about 30 minutes either side of each unsafe window. |
| 6 (criterion C) | The note gives the times when boats cannot dock (08:59 to 14:01), the arrivals that can dock, the margin, and a warning that the later timetable moves with the tide, with units, in sentences that make sense without the working. |
Exemplar answers
| Task D1 response | What it looks like |
|---|---|
| Level 3 to 4 (D) | “Greatest depth 3.8, least 1.4. At 2 hours the depth is 3.2. Boats cannot dock around t = 4. The 10:00 boat cannot dock.” Correct values, but the interval for docking is not found, the arrivals are not all checked and nothing is said about the model. |
| Level 7 to 8 (D and C) | “2.6 + 1.2 cos 30t = 2.3 gives cos 30t = −0.25, so 30t = 104.5° or 255.5°, t = 3.48 or 8.52 hours. No docking from 08:59 to 14:01. 10:00, 11:30 and 13:00 are inside, the other four are outside. A real cycle of 12 h 25 min stretches every time after high tide by about 3.5%, so the unsafe window becomes about 09:06 to 14:19 and the next high tide is near 17:55, not 17:30. I recommend treating 08:30 and 14:30 as borderline and adding 30 minutes either side.” The note is five clear sentences with units and clock times. |
Differentiation
A way in for anyone stuck
- Give a copy of the cosine graph from lesson 2 with the line d = 2.3 drawn on it so that the two crossings can be read before they are calculated.
- Allow the time to be converted to minutes on a number line from 0 to 60.
- For Part 3, ask: “If the next high tide came 25 minutes later, what would happen to the times you found?”
Stretch
- The operator also wants to know how many hours in a whole day the pier cannot be used if the model repeats exactly. (Two intervals of 5.03 hours: 10.1 hours.)
- Write a model of the depth with t measured from midnight instead of from high tide. (d = 2.6 + 1.2 cos 30(t − 5.5).)
Students learning in English as an additional language
| Word in task D1 | Plain meaning |
|---|---|
| depth | how deep the water is |
| high tide | the time when the water is at its highest |
| dock | bring a boat to the pier to load and unload |
| at least | equal to or more than |
| recommend | say what someone should do, and why |
| margin | extra time or space left for safety |
Sentence starters for task D1:
- The depth is at least 2.3 m when …
- The boat can dock at … because the depth is …
- I recommend … because the model …
Task D2: Two boats leave the harbour
Why set it. The cosine rule, the sine rule and bearings used together in one decision about radio contact, with a test of how far the answer can be trusted. An alternative to D1: set one of them.
When: End of unit, after lesson 6. Time: 50 minutes. Format: Individual, in class, one lesson. A written solution with explanations and a short note to the harbour radio operator.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 6. Time: 50 minutes. Calculator: scientific calculator. Criteria: D, applying mathematics in real-life contexts; C, communicating.
Two fishing boats leave a harbour H on the Andaman coast at the same time. Boat A sails at 24 km/h on a bearing of 070°. Boat B sails at 18 km/h on a bearing of 145°. Both keep a steady speed and course. Their radios work only when the boats are 15 km or less apart.
Part 1. After 30 minutes
- Draw a labelled diagram of the harbour and the two boats after 30 minutes, and find the angle AHB between their paths.
- Find the distance between the boats after 30 minutes.
- Find the bearing of boat B from boat A after 30 minutes, to the nearest degree.
Part 2. Losing contact
- After how many minutes do the boats lose radio contact? Show how you found it, and give your answer to the nearest minute.
Part 3. Does it make sense?
- The bearings are measured to the nearest degree. Find how much the distance in question 2 could change if the bearing of boat B were really 143° or 147°, and say what that means for your answer to question 4.
- Write a short note to the harbour radio operator, saying when to expect radio contact to end and how sure you are, so that someone who has not seen your working can understand it.
How your work is judged
| Level | Criterion D in task D2: what it looks like |
|---|---|
| 1–2 | You find some of the right numbers, for example the angle AHB or a distance travelled, but there is no answer about radio contact. |
| 3–4 | You find the distance and the bearing after 30 minutes and a time for losing contact, and say whether your answers make sense. |
| 5–6 | You choose suitable methods throughout (angles, the cosine rule, the sine rule), reach correct answers, explain how accurate the time is and whether the answer is sensible in practice. |
| 7–8 | You choose and use suitable methods throughout, reach correct answers, explain how accurate your time is and why, and justify whether your recommendation is reasonable for real boats. |
| Level | Criterion C in task D2: what it looks like |
|---|---|
| 1–2 | Some working is shown, but it is hard to see how you reached the distance, the bearing or the time. |
| 3–4 | Your working for the distance and bearing can be followed, with some correct notation, units and a diagram. |
| 5–6 | Your working is clear and in order, with a labelled diagram, usually correct units and bearings written with three figures, and your note is mostly well organized. |
| 7–8 | Your working is clear, concise and well ordered, your diagram matches your calculations, notation and units are always correct, and your note makes sense on its own. |
Markscheme
| Task D2 question | Answer and marking |
|---|---|
| 1 | Angle AHB = 75° (145° − 70°), with distances HA = 12 km and HB = 9 km after 30 minutes (24 × 0.5 and 18 × 0.5). Credit a diagram with H, A and B marked, the bearings measured clockwise from north, and the lengths. |
| 2 | 13.0 km. AB2 = 122 + 92 − 2 × 12 × 9 × cos 75° = 225 − 55.9 = 169.1, AB = 13.00 km. 1 mark for the cosine rule set up, 1 for the value 169.1, 1 for 13.0. Leaving out the cosine term gives exactly 15 km, which is also the radio limit and an easy trap. |
| 3 | 208°. sin HAB = 9 sin 75° ÷ 13.00 = 0.669, angle HAB = 42.0°. The bearing of H from A is 070° + 180° = 250°, and B lies on the side of the line AH towards the south, so the bearing of B from A is 250° − 42.0° = 208.0°, 208° to the nearest degree. 1 mark for the sine rule, 1 for 42.0°, 1 for the bearing. 292° adds the angle instead of subtracting it. |
| 4 | 35 minutes. After t hours the sides are 24t and 18t, so AB2 = (24t)2 + (18t)2 − 2 × 24t × 18t × cos 75° = 676.38t2, AB = 26.01t km. Setting 26.01t = 15 gives t = 0.577 hours = 34.6 minutes, 35 minutes. 1 mark for expressing AB in terms of t, 1 for the equation, 1 for the time. Credit a proportion argument: AB is 13.00 km at 0.5 hours and grows steadily, so 15 km comes at 0.5 × 15 ÷ 13.00 hours. |
| 5 | An angle of 73° gives AB = 12.72 km after 30 minutes and 77° gives 13.28 km, so the distance changes by about 0.3 km. The time to lose contact is 450 ÷ 12.72 = 35.4 minutes and 450 ÷ 13.28 = 33.9 minutes, within about a minute either way of 35. The answer should be quoted to the nearest minute at most. Credit any reasoned discussion with units; credit the real-life point that boats do not hold a perfectly steady speed or course. |
| 6 (criterion C) | The note gives the time to expect contact to end (about 35 minutes after leaving), the distance and bearing at 30 minutes, the sensitivity (about a minute) and a warning about steady courses, with units, in sentences that make sense without the working. |
Exemplar answers
| Task D2 response | What it looks like |
|---|---|
| Level 3 to 4 (D) | “Angle is 75. The boats are 12 km and 9 km out after 30 minutes. Using the cosine rule they are 13.0 km apart. They are in contact. The bearing of B is about 208.” Correct values, but no time for losing contact is found and nothing is said about accuracy. |
| Level 7 to 8 (D and C) | “AB = 26.01t km, so contact is lost when 26.01t = 15, t = 0.577 h = 35 minutes. If B’s bearing is 143° or 147°, the distance at 30 minutes is 12.7 km or 13.3 km, so contact is lost about a minute later or earlier. I recommend telling the crews to expect contact to end 34 to 36 minutes after leaving, as long as both keep a steady course and speed.” Labelled diagram, bearings in three figures, units everywhere, and a five-sentence note. |
Differentiation
A way in for anyone stuck
- Give a pre-drawn diagram of the harbour with north lines at each point, so the angle at A can be placed.
- Allow the 30 minute distances to be found first, before the general time t is introduced.
- For Part 3, ask: “If the angle between the boats were a little bigger, would they be further apart or closer?”
Stretch
- Find the distance between the boats when boat A has been sailing for 45 minutes and boat B for 30 minutes. (17.9 km.)
- Find the area of triangle HAB after 30 minutes and use it to find the shortest distance from the harbour to the line joining the boats. (52.2 km2; 8.0 km.)
Students learning in English as an additional language
| Word in task D2 | Plain meaning |
|---|---|
| bearing | a direction measured clockwise from north, written with three figures such as 070° |
| steady | not changing |
| radio contact | the boats can hear each other on the radio |
| at the same time | starting together |
| sensible | reasonable and likely to be true in real life |
Sentence starters for task D2:
- After 30 minutes the boats are … km from the harbour, so …
- The bearing of B from A is … because …
- I am sure to within … because …
Practicalities
Every task is original and written for this unit; none is copied or adapted from a published book.
The student sheets, with their task-specific clarifications, are public on the unit page so students know what each level looks like before they start. Markschemes and exemplars are only on this page.
Calculators: every task needs a scientific calculator set to degrees, except the exact-value questions (A1 question 1, A2 question 1), where it should be put away. A GDC is not needed.
The sheets are public, so a student may have seen A1 and A2. For an unseen A1, use this parallel version (same marks, same markscheme notes): Q1 (a) sin 300° (b) tan 150° (−√3 ÷ 2; −1 ÷ √3). Q2 given sin 40° = 0.6428, the values of sin 140°, sin 220° and sin 320° (0.6428; −0.6428; −0.6428). Q3 y = 2 sin 3x − 1: amplitude, period, least value (2; 120°; −3). Q4 y = a sin bx has greatest value 4 and first returns to the x-axis at x = 90° (a = 4, b = 2). Q5 cos x = 0.35 (69.5°, 290.5°). Q6 3 sin x − 1 = −2.5 (210°, 330°). Q7 3 sin x = 4 cos x (53.1°, 233.1°). Q8 the number of solutions of sin x = 0.6 for 0° ≤ x ≤ 720° (4).
For an unseen A2, use this parallel version (same marks, same markscheme notes): Q1 (a) cos 240° (b) sin 315° (−1 ÷ 2; −√2 ÷ 2). Q2 y = 4 cos 2x: amplitude, period, number of waves from 0° to 360° (4; 180°; 2). Q3 7 cos x = 3 (64.6°, 295.4°). Q4 A = 52°, B = 67°, a = 8.4 cm, find b (9.81 cm). Q5 a wheel with h = 11 − 10 cos 36t, t in minutes: least height 1 m, greatest 21 m, period 10 minutes, first time at 15 m is 3.2 minutes. Q6 a triangle with sides 8, 11 and 14 cm: the largest angle (93.6°) and the area (43.9 cm2). Q7 X = 30°, XY = 10 cm, YZ = 6 cm: Z = 56.4° or 123.6°. Q8 cos x = sin 20° for 0° ≤ x ≤ 360° (70°, 290°). Q9 sides 6 cm and 10 cm and area 21 cm2: included angle 44.4° or 135.6°.
The Year 5 descriptor bands on this page are band summaries written for this site; check them against the current IB MYP Mathematics guide before relying on them for reporting.