Right-angled triangles: a week of ready-to-teach maths
Five lessons for Year 9 Measurement. Print this pack and the week is prepared: each day has a one-page plan and a student worksheet with practice and a stretch section, plus four triangles to measure, which-tool cards, a clinometer to make, a mini-check and every answer with its working.
Start here: five minutes to Monday
- Skim the week at a glance on the next page.
- Print the five days. Each day is two A4 sheets: a plan and a worksheet.
- Print the two cut-out sheets, one copy per pair: sheet 1 has the ratio triangles (Day 4) and the which-tool cards (Day 5), sheet 2 a clinometer (Day 5). Cut the cards before Day 5.
- Calculators with sin, cos and tan are needed from Day 3; rulers on Days 2 and 4; for the clinometer, a straw, string and a small weight per pair.
- Open the free interactive unit on the board. Every plan says which picture to show and which buttons to press.
The idea in plain words
In a right-angled triangle the two acute angles add to 90°, so one fixes the other. Triangles with the same angles are similar: every side is multiplied by the same scale factor, which is how a shadow measures a tree. Pythagoras’ theorem links the three sides, c² = a² + b², with c the hypotenuse opposite the right angle. For a chosen acute angle, the ratios of the sides are fixed whatever the size of the triangle: sine = opposite ÷ hypotenuse, cosine = adjacent ÷ hypotenuse, tangent = opposite ÷ adjacent. So an angle and one side give every other side, and two sides give the angle.
One day, one lesson
The five lessons fill a week of maths, one lesson of about 60 minutes a day. Run them in order: each day stands on the one before. Every plan marks a split point for two shorter sessions, and every worksheet ends with a stretch section for students who finish early.
How we write these answers
Lengths are rounded to two decimal places and angles to one, unless a question says otherwise; an irrational length is given exactly first, as a surd such as √74 or 2√13. Calculators must be in degree mode. Every working names the ratio it uses, for example opposite = 20 × sin 25°.
The week at a glance
One lesson a day for a week. Each day stands on the day before, so run them in order.
| Day | Lesson | Students learn and do | On screen |
|---|---|---|---|
| 1 | Angles and names | Use the angle sum to find the second acute angle, name the hypotenuse, opposite and adjacent for a chosen angle, and solve angle problems about ramps, ladders and shadows | Naming the sides; Angles add to 180, the acute pair to 90 |
| 2 | Similar triangles and scale | Find scale factors and missing sides of similar right triangles, measure heights from shadows, and read a right-angled plan at 1 : 200 | Similar triangles and scale factor |
| 3 | Pythagoras | Find a hypotenuse and a shorter side, give an irrational answer exactly and to two decimal places, and test whether three sides make a right angle | Pythagoras: squares on the sides; An irrational hypotenuse |
| 4 | The trigonometric ratios | Measure four 35° triangles to see that each ratio stays fixed, write sin, cos and tan as fractions, and find a side with the right ratio | Trig finds a side |
| 5 | Solving spatial problems | Find angles with inverse ratios, measure a height with a clinometer, choose between Pythagoras, trigonometry and similar triangles, and play the quiz as a class game | Self-check quiz as a class game |
How the week builds
Day 1 sets the angle facts and the names of the sides; Day 2 uses similar triangles and scale; Day 3 is Pythagoras, with exact and rounded answers; Day 4 builds the three ratios from measured triangles; and Day 5 finds angles and chooses the right tool for a spatial problem. It builds on Year 8 work with Pythagoras’ theorem and similar triangles, and on surds earlier in Year 9. It prepares Year 10 problems with directions and angles of elevation and depression.
Materials for the week (one trip)
- From the classroom: calculators with sin, cos and tan, rulers, protractors, scissors, sticky tape, envelopes for the cards, this pack printed.
- Cut-out sheet 1, one per pair: four ratio triangles (Day 4) and 12 which-tool cards (Day 5).
- Cut-out sheet 2, one per pair: a clinometer (Day 5), with a drinking straw, about 30 cm of string and a washer or paper clip.
- If it is sunny: a metre stick for measuring shadows on Day 2. No other equipment to buy.
Words for the word wall
right angle, hypotenuse, opposite, adjacent, acute, complementary, angle sum, similar, scale factor, enlargement, Pythagoras’ theorem, surd, irrational, sine, cosine, tangent, ratio, inverse, clinometer, angle up from level
Dear families
This week in maths, Year 9 is solving problems with right-angled triangles: finding a missing side with Pythagoras’ theorem, using shadows and similar triangles to measure heights, and using sine, cosine and tangent to link angles and sides. On Friday students make a clinometer to measure the height of something tall.
Try this at home
- Measure a rectangular table or door, then predict its diagonal with Pythagoras’ theorem and check with a tape measure.
- Ask: a 3 m ladder stands with its foot 1 m from a wall. How high does it reach? (About 2.83 m: √8.)
- Ask: on a sunny day, why do a person and a tree have shadows in the same ratio to their heights? (The Sun’s rays are parallel.)
- Ask: what does SOH CAH TOA stand for? (Sine = opposite ÷ hypotenuse, and so on.)
My triangle week
Fill one row a day. Draw or describe a triangle from class, what was missing, and how you found it.
| Day | Triangle | Missing | How I found it |
|---|---|---|---|
| Monday | |||
| Tuesday | |||
| Wednesday | |||
| Thursday | |||
| Friday |
Angles and names
The right angle takes 90° of a triangle’s 180°, so the two acute angles share the other 90°: one gives the other. The sides are named from a chosen acute angle: the hypotenuse faces the right angle, the opposite faces the chosen angle, and the adjacent touches it.
We are learning to
- find the second acute angle of a right-angled triangle from the first,
- name the hypotenuse, opposite and adjacent for either acute angle,
- use angle facts in problems about ramps, ladders and shadows.
Success criteria
- I can explain why the two acute angles of a right-angled triangle add to 90°.
- I can name the opposite and adjacent sides for each acute angle of the same triangle.
You need
The worksheet; protractors.
Lesson flow (about 60 minutes)
| 12 min | One angle gives the other Show the first board picture (see the box). Then a right-angled triangle with one acute angle of 28°: the other is 90° − 28° = 62°. Check with a protractor on a sketch. Ask: “Could a right-angled triangle have an angle of 95°? Why not?” |
| 13 min | Which side is which? Show the second picture, then pairs do Part B: they name every side for one angle, then for the other. Ask: “Which name never changes when you switch angles?” |
| 35 min | Angles at work Parts A and C; early finishers do Part D. Ask for a sketch with the right angle marked for each Part C problem. Ask: “Where is the right angle in the ladder problem?” |
Two shorter sessions instead? Split after Part B; start the second with the board pictures again.
Watch for these ideas
- Naming the opposite and adjacent before choosing the angle: they depend on it.
- Calling the side opposite the chosen angle the hypotenuse: the hypotenuse faces the right angle.
- Subtracting from 180° and forgetting the right angle: 180 − 90 − the angle.
Answers
- Part A: 41° gives 49°; 17° gives 73°; 63.5° gives 26.5°; 45° gives 45°.
- Part B: at P, hypotenuse PR, opposite QR, adjacent PQ; at R, hypotenuse PR, opposite PQ, adjacent QR. At L, hypotenuse KL, opposite KM, adjacent LM.
- Part C: 85°; 18°; 52°: each time the two acute angles of a right-angled triangle add to 90°.
- Part D: no and no: an obtuse angle or a second right angle with the first would already reach more than 180° or leave 0°.
Angles and names
A. The other acute angle
Each angle is one acute angle of a right-angled triangle. Find the other.
| 41° and ______ | 17° and ______ | 63.5° and ______ | 45° and ______ |
B. Name the sides
Name each side by its two letters.
| Angle | hypotenuse | opposite | adjacent |
|---|---|---|---|
| the marked angle at P | |||
| the angle at R | |||
| the marked angle at L |
C. Angles at work
Sketch each one with its right angle marked.
| A ramp meets level ground at 5°. The wall at its top is vertical. What angle does the ramp make with the wall? | |
| A ladder leans against a wall at 72° to the ground. What angle does it make with the wall? | |
| The Sun’s rays reach the ground at 38°. What angle do they make with a vertical pole? |
D. Stretch: impossible triangles
Can a right-angled triangle have an obtuse angle? Can it have two right angles? Explain with the angle sum.
Similar triangles and scale
Triangles with the same angles are similar: one is an enlargement of the other, every side multiplied by the same scale factor. Shadows cast at the same moment make similar triangles, so a short stick can measure a tall tree.
We are learning to
- find a scale factor from one pair of matching sides, and use it for the others,
- measure a height from shadows with similar triangles,
- turn lengths on a scale plan into real lengths.
Success criteria
- I can find the scale factor between two similar triangles and use it to find a missing side.
- I can explain why a stick and a tree make similar triangles with their shadows.
You need
The worksheet; rulers; calculators; a metre stick and the playground if it is sunny.
Lesson flow (about 60 minutes)
| 12 min | Same angles, scaled sides Show the board (see the box). Then shadows: a 1.5 m stick casts a 2 m shadow while a building’s shadow is 24 m. The scale factor is 24 ÷ 2 = 12, so the building is 1.5 × 12 = 18 m tall. Ask: “Why must the two shadows be measured at the same time?” |
| 13 min | Scale factors Pairs do Part A: match the given side to its partner first, then find the factor. Ask: “Can a scale factor be less than 1? What does the copy look like then?” |
| 35 min | Shadows and plans Parts B and C; early finishers do Part D. If it is sunny, pairs measure a metre stick’s shadow and the shadow of something tall outside, and work out its height. Ask: “Which measurement would you most like to take twice, and why?” |
Two shorter sessions instead? Split after Part A; start the second outside with the shadows.
Watch for these ideas
- Adding instead of multiplying: 3, 4, 5 enlarged is not 6, 7, 8.
- Matching the wrong sides: shortest goes with shortest, hypotenuse with hypotenuse.
- Dividing when the copy is bigger: a factor above 1 makes every side longer.
Answers
- Part A: × 3: 9, 12, 15; × 0.5: 2.5, 6, 6.5; × 1.5: 12, 22.5, 25.5; × 2: 14, 48, 50.
- Part B: a flagpole: 6 ÷ 0.9 × 1.2 = 8 m; a gum tree: 15 ÷ 2.4 × 1.6 = 10 m.
- Part C: 1 cm on the plan is 200 cm = 2 m, so the legs are 6 m and 8 m and the slanting side is 5 cm on the plan, 10 m for real.
- Part D: the Sun’s rays are parallel, so they meet the level ground at the same angle; the stick and the pole are both vertical, so each triangle has a right angle. Same angles, so similar. At different times the ray angle changes and the triangles are not similar.
Similar triangles and scale
A. Scale factors
Each right-angled triangle is enlarged or reduced. Find the scale factor, then the missing sides.
| Triangle | Copy | Scale factor | Missing sides |
|---|---|---|---|
| 3, 4, 5 | ___, 12, ___ | ||
| 5, 12, 13 | ___, ___, 6.5 | ||
| 8, 15, 17 | 12, ___, ___ | ||
| 7, 24, 25 | ___, 48, ___ |
B. Shadows
Each pair of shadows was measured at the same time. Sketch the two triangles, then find the height.
| The shadow of a 1.2 m stick is 0.9 m long. The shadow of a flagpole is 6 m. How tall is the flagpole? | |
| The shadow of a student 1.6 m tall is 2.4 m long. The shadow of a gum tree is 15 m. How tall is the gum tree? |
C. A plan at 1 : 200
A right-angled garden bed is drawn on a plan at 1 : 200. Its two shorter sides measure 3 cm and 4 cm on the plan. Measure or work out its slanting side on the plan, then give all three real lengths in metres.
D. Stretch: why similar?
Explain why a stick and a flagpole make similar triangles with their shadows. Why must both shadows be measured at the same time?
Pythagoras
In a right-angled triangle the square on the hypotenuse equals the squares on the two shorter sides together: c² = a² + b². Adding finds the hypotenuse and subtracting finds a shorter side. Most answers are irrational: give the surd, then the decimal.
We are learning to
- find the hypotenuse from the two shorter sides,
- find a shorter side from the hypotenuse and the other side,
- give an irrational answer exactly and rounded, and test whether three sides make a right angle.
Success criteria
- I can find a missing side and give it exactly and to two decimal places.
- I can tell from its three sides whether a triangle is right-angled.
You need
The worksheet; calculators; grid paper for anyone who wants to draw the squares.
Lesson flow (about 60 minutes)
| 12 min | Squares on the sides Show the first board picture (see the box). Then shorter sides 9 and 12: c² = 81 + 144 = 225, so c = 15. Ask: “Why add the squares and not the sides? What would 9 + 12 give?” |
| 13 min | Exact, then rounded Show the second picture, then pairs do Part A, writing the surd before the decimal. Ask: “Why is √2 ≈ 1.41 and not √2 = 1.41?” |
| 35 min | Backwards and the test Parts B and C; early finishers do Part D. Ask: “To find a shorter side, which square do you start from? Why subtract?” |
Two shorter sessions instead? Split after Part A; start the second with the squares on the board again.
Watch for these ideas
- Adding the sides, not their squares: shorter sides 9 and 12 give 15, not 21.
- Adding to find a shorter side: subtract the known square from the square on the hypotenuse.
- Rounding too early, or writing √74 = 8.60: the decimal is only close.
Answers
- Part A: 64 + 225 = 289, c = 17; 400 + 441 = 841, c = 29; 25 + 49 = 74, c = √74 ≈ 8.60; 6.25 + 36 = 42.25, c = 6.5.
- Part B: 625 − 49 = 576, side = 24; 1681 − 81 = 1600, side = 40; 144 − 25 = 119, side = √119 ≈ 10.91.
- Part C: 81 + 1600 = 1681 = 1681: yes; 2.25 + 4 = 6.25 = 6.25: yes; 36 + 81 = 117 ≠ 121: no; 100 + 576 = 676 ≠ 625: no.
- Part D: the diagonal is √50 = √(25 × 2) = 5√2 ≈ 7.07 cm.
Pythagoras
A. Find the hypotenuse
Give each answer exactly, and to two decimal places if it is not a whole number or a short decimal.
| Shorter sides | Working | Hypotenuse |
|---|---|---|
| 8 and 15 | ||
| 20 and 21 | ||
| 5 and 7 | ||
| 2.5 and 6 |
B. Find a shorter side
| Hypotenuse, side | Working | Other side |
|---|---|---|
| 25, 7 | ||
| 41, 9 | ||
| 12, 5 |
C. Right-angled or not?
Square the two shorter sides and add. Is the total the square of the longest side?
| 9, 40, 41 | |
| 1.5, 2, 2.5 | |
| 6, 9, 11 | |
| 10, 24, 25 |
D. Stretch: the diagonal of a square
A square has sides of 5 cm. Find its diagonal exactly, as a simplified surd, and to two decimal places.
The trigonometric ratios
For a chosen acute angle, each ratio of two sides is the same in every right-angled triangle with that angle, because those triangles are similar. Sine, cosine and tangent name the three ratios, so an angle and one side give any other side.
We are learning to
- measure the ratios of sides in similar triangles and see that they stay fixed,
- write sin θ, cos θ and tan θ as fractions of the sides,
- choose the ratio that links two sides and use it to find a side.
Success criteria
- I can write sin θ, cos θ and tan θ for a triangle with sides 3, 4 and 5.
- I can choose the ratio that links the side I know with the side I want.
You need
The worksheet; rulers; calculators in degree mode; the ratio triangles (cut-out sheet 1), one set per pair.
Lesson flow (about 60 minutes)
| 15 min | Measure four triangles Pairs measure triangles A to D in millimetres and divide (Part C). Every triangle gives about 0.57, 0.82 and 0.70: then they press sin 35, cos 35 and tan 35 on the calculator. Ask: “The triangles are different sizes. Why are the ratios the same?” |
| 10 min | Trig finds a side Show the board (see the box). Then an angle of 40° with a hypotenuse of 12 cm: opposite = 12 × sin 40° ≈ 7.71 cm and adjacent = 12 × cos 40° ≈ 9.19 cm. Ask: “Which ratio links the side you know with the side you want?” |
| 35 min | Ratios and sides Parts A and B; early finishers do Part D. Ask for the ratio and its working before the answer. Ask: “When the side you want is the hypotenuse, why do you divide?” |
Two shorter sessions instead? Split after measuring; start the second with the board picture.
Watch for these ideas
- A calculator in radian mode: sin 30 should give 0.5.
- Multiplying when the unknown is the hypotenuse: hypotenuse = opposite ÷ sin θ.
- Thinking a bigger triangle has a bigger sine: the ratio depends only on the angle.
Answers
- Part A: triangle 1: sin θ = 5/13, cos θ = 12/13, tan θ = 5/12; triangle 2: sin θ = 15/17, cos θ = 8/17, tan θ = 15/8.
- Part B: opposite = 20 × sin 25° ≈ 8.45 cm; opposite = 6 × tan 52° ≈ 7.68 cm; adjacent = 9 × cos 18° ≈ 8.56 cm; hypotenuse = 10 ÷ sin 63° ≈ 11.22 cm.
- Part C: about 0.57, 0.82 and 0.70 for every triangle (sin 35° = 0.5736, cos 35° = 0.8192, tan 35° = 0.7002); a difference in the second decimal place is measuring error.
- Part D: the hypotenuse is the longest side, so opposite ÷ hypotenuse is less than 1. Tangent is opposite ÷ adjacent, which is more than 1 once the angle passes 45° and the opposite is longer than the adjacent.
The trigonometric ratios
A. Write the ratios
For the marked angle θ, write sin θ, cos θ and tan θ as fractions.
cos θ =
tan θ =
cos θ =
tan θ =
B. Find the side
Name the ratio that links the two sides, then find the side to two decimal places. Lengths are in cm.
| Given | Find | Ratio and working |
|---|---|---|
| angle 25°, hypotenuse 20 | opposite | |
| angle 52°, adjacent 6 | opposite | |
| angle 18°, hypotenuse 9 | adjacent | |
| angle 63°, opposite 10 | hypotenuse |
C. Measure the ratio triangles
Each triangle on cut-out sheet 1 has a 35° angle. Measure its sides in mm, then divide to two decimal places.
| Triangle | opp | adj | hyp | opp ÷ hyp | adj ÷ hyp | opp ÷ adj |
|---|---|---|---|---|---|---|
| A | ||||||
| B | ||||||
| C | ||||||
| D |
D. Stretch: how big can they be?
Explain why sin θ is never more than 1, but tan θ can be. For which angles is tan θ more than 1?
Solving spatial problems
Two sides give an angle through an inverse ratio: if tan θ = 0.75, then θ = tan⁻¹ 0.75. A spatial problem starts with a sketch and its right angle; the facts given then choose the tool: two sides for Pythagoras, an angle and a side for trigonometry, matching shapes for similar triangles.
We are learning to
- find an angle from two sides with sin⁻¹, cos⁻¹ or tan⁻¹,
- measure a height with a clinometer, a distance and some trigonometry,
- choose Pythagoras, trigonometry or similar triangles to fit a problem.
Success criteria
- I can find an angle from two sides with the inverse of the right ratio.
- I can say which tool a problem needs before I calculate, and why.
You need
The worksheet; calculators; the clinometer (cut-out sheet 2) with a straw, string and a washer; the which-tool cards (cut-out sheet 1); the unit’s self-check quiz. Never sight the Sun through the straw.
Lesson flow (about 60 minutes)
| 10 min | The ladder A 4 m ladder with its foot 1.2 m from a wall: it reaches √(16 − 1.44) ≈ 3.82 m up (Pythagoras) and makes cos⁻¹(1.2 ÷ 4) ≈ 72.5° with the ground (trigonometry). Ask: “Which tool did each step need, and why?” |
| 20 min | Measure a height Pairs make the clinometer and measure something tall outside: pace or tape the distance, read the angle, then height = distance × tan(angle) + eye height. Ask: “Why add the eye height? What if you stood twice as far away?” |
| 15 min | Angles and problems Parts A and B, a sketch with the right angle marked for every problem. |
| 15 min | Which tool? and the quiz Pairs sort the which-tool cards (Part C). Then run the unit’s self-check quiz as a class game: teams vote, then the reason is read aloud. |
Two shorter sessions instead? Split after measuring the height; start the second with Part B. Early finishers do Part D.
Watch for these ideas
- Pressing sin instead of sin⁻¹ to find an angle: the answer should be in degrees, not a ratio.
- Forgetting the eye height, or measuring the distance to the wrong point.
- Reaching for trigonometry when two sides and no angle are given: that is Pythagoras.
Answers
- Part A: sin⁻¹(5 ÷ 9) ≈ 33.7°; cos⁻¹(7 ÷ 10) ≈ 45.6°; tan⁻¹(11 ÷ 6) ≈ 61.4°.
- Part B: 1. tan⁻¹(0.5 ÷ 7) ≈ 4.1°, length √(49 + 0.25) ≈ 7.02 m. 2. 30 × sin 50° ≈ 22.98 m. 3. 15 × tan 38° + 1.5 ≈ 13.22 m. 4. Across 130 m, around 170 m: 40 m shorter.
- Part C: the card key is with the mini-check answers.
- Part D: 25 × tan 32° + 1.6 ≈ 17.22 m; from 40 m, tan⁻¹(15.62 ÷ 40) ≈ 21.3°.
Solving spatial problems
A. Find the angle
Name the ratio, then use its inverse. Give the angle to one decimal place.
| opposite 5, hypotenuse 9 | θ = |
| adjacent 7, hypotenuse 10 | θ = |
| opposite 11, adjacent 6 | θ = |
B. Spatial problems
Sketch each one and mark the right angle. Then name the tool and solve.
| 1. An access ramp rises 0.5 m over 7 m along level ground. At what angle does it rise, and how long is the ramp? | |
| 2. A kite string 30 m long is pulled straight and makes 50° with the ground. How high is the kite above the hand holding it? | |
| 3. Standing 15 m from a tree, a student sights its top through a clinometer at 38°. Their eyes are 1.5 m above the ground. How tall is the tree? | |
| 4. A park is a rectangle 120 m by 50 m. How much shorter is the walk straight across, corner to corner, than along two sides? |
My clinometer height: object ____________ distance ______ m angle ______° eye height ______ m height ≈ ______ m
C. Which tool?
Sort the which-tool cards. Write each card’s letter under the tool it needs.
| Pythagoras | trigonometry | similar triangles |
|---|---|---|
D. Stretch: step back
From 25 m away, the top of a building is 32° up through a clinometer held 1.6 m above the ground. How tall is the building? What angle would you read from 40 m away?
Ratio triangles and which-tool cards
Ratio triangles (Day 4), one set per pair
Each triangle has a 35° angle. Measure its three sides in millimetres; you do not need to cut these out.
Which-tool cards (Day 5), one set per pair: 12 cards
Sort the cards: does each problem need Pythagoras, trigonometry or similar triangles?
Teacher note: the card key is with the mini-check answers, so this sheet can go to pairs whole.
Make a clinometer
For Day 5, one per pair
- Glue the sheet to thin card if you can, and cut out the quarter circle.
- Tape a drinking straw along the top edge, and push the string through the hole at the corner. Tie a washer or paper clip to the end.
- Hold the corner end to your eye and look along the straw at the top of something tall. Never look at the Sun.
- Keep still until the string hangs at rest, pinch it against the card, and read the angle where it crosses the scale.
Teacher note: the scale reads the angle up from level directly (0° when the straw is level), so height = distance × tan(angle) + eye height.
What we know: Right-angled triangles
Work on your own. Use a calculator in degree mode. Give lengths to two decimal places and angles to one.
- One acute angle of a right-angled triangle is 34°. What is the other? ______________________
- In triangle XYZ the right angle is at Y. For the angle at X, which side is the opposite, and which is the adjacent? ______________________
- A person 1.8 m tall casts a 2.7 m shadow. At the same moment a tower’s shadow is 36 m. How tall is the tower? ______________________
- The two shorter sides of a right-angled triangle are 12 cm and 35 cm. Find the hypotenuse. ______________________
- The hypotenuse is 26 cm and one shorter side is 10 cm. Find the other side. ______________________
- The shorter sides are 4 cm and 6 cm. Find the hypotenuse exactly, as a simplified surd, and to two decimal places. ______________________
- A right-angled triangle has a 48° angle and a hypotenuse of 15 cm. Find the side opposite the 48° angle. ______________________
- The side opposite an angle is 7 cm and the side adjacent to it is 12 cm. Find the angle. ______________________
How did this week feel? Circle one: I can teach it · I am getting there · I need more help
Answers and marking guide
Answers with working
- 90° − 34° = 56°.
- The opposite faces X: YZ. The adjacent touches X and is not the hypotenuse: XY. (The hypotenuse is XZ.)
- Scale factor 36 ÷ 2.7 = 13.333333, so the tower is 1.8 × 13.333333 = 24 m.
- 144 + 1225 = 1369, so c = 37 cm.
- 676 − 100 = 576, so the side is 24 cm.
- 16 + 36 = 52, so c = √52 = 2√13 ≈ 7.21 cm.
- Opposite and hypotenuse: sine. 15 × sin 48° ≈ 11.15 cm.
- Opposite and adjacent: tangent. tan⁻¹(7 ÷ 12) ≈ 30.3°.
A quick three-level guide
| Idea | Working towards | At standard | Beyond |
|---|---|---|---|
| Angles and names | Subtracts from 180° only, or names sides before choosing the angle | Finds the other acute angle and names hypotenuse, opposite and adjacent | Explains why the acute angles of a right-angled triangle add to 90° |
| Similarity and scale | Adds instead of multiplying, or matches the wrong sides | Finds a scale factor and uses it for missing sides and heights | Explains why shadows at the same moment make similar triangles |
| Pythagoras | Adds the sides, or adds when a shorter side is wanted | Finds any side and gives irrational answers exactly and rounded | Tests for a right angle and simplifies surds |
| Trigonometry | Picks a ratio at random, or uses sin for sin⁻¹ | Chooses the ratio for the sides involved and finds a side or an angle | Plans a spatial problem, chooses the tool and checks it is sensible |
Eight questions, four ideas. A student at standard answers most questions with working, naming the ratio or the theorem each time.
Card key
Which tool (Day 5): Pythagoras: B, D, E, I, K; trigonometry: A, F, G, L; similar triangles: C, H, J.
Weekly class record
Jot a tick as you move around the room; the mini-check fills any gaps. A tick a day is plenty.
| Name | Angles | Similar | Pythagoras | Ratios | Problems |
|---|---|---|---|---|---|
The five columns are the five days: angles and names, similar triangles, Pythagoras, the ratios, and spatial problems.