[{"data":1,"prerenderedAt":1107},["ShallowReactive",2],{"layer:constructing-angles:extend":3},{"layer":4,"contentHash":1079,"dependencyHashes":1080,"approval":1101,"releaseId":1106},{"schemaVersion":5,"conceptId":6,"locale":7,"depth":8,"revision":5,"title":9,"subtitle":10,"summary":11,"objectives":12,"estimatedMinutes":18,"plate":19,"blocks":40,"sourceIds":1074,"reviewStatus":1075,"authoring":1076},1,"constructing-angles","en","extend","Triangles, polygons and the impossible angle","Build triangles and regular polygons, meet Gauss's 17-gon, and find out why 20° can never be constructed","Construct triangles from SSS, SAS and ASA, draw regular polygons from a circle, discover which polygons and whole-degree angles are constructible (multiples of 3°), meet the trisection problem, and use angles in projects, puzzles and careers.",[13,14,15,16,17],"Construct triangles from SSS, SAS and ASA data, check them with the triangle inequality and a protractor, and explain why SSA and AAA do not fix a triangle.","Construct regular polygons from a circle and calculate central, interior and mitre angles.","Explain which regular polygons and which whole-degree angles can be constructed, and why 20° cannot.","Use constructed angles in projects such as a clinometer and in olympiad-style puzzles about clocks and bisections.","Describe how carpenters, architects, surveyors and engineers use angles today.",50,{"title":20,"rows":21},"Lesson plate",[22,25,28,31,34,37],{"label":23,"value":24},"Depth","Extend",{"label":26,"value":27},"Reading time","≈ 50 minutes",{"label":29,"value":30},"Prior knowledge","All constructions (Understand, Deepen)",{"label":32,"value":33},"Chapters","10",{"label":35,"value":36},"Labs","Criteria match, constructible sort, 3 more",{"label":38,"value":39},"Big surprise","20° is impossible",[41,45,51,57,60,85,99,109,119,124,129,164,179,184,187,239,255,262,267,280,283,301,307,312,315,320,353,364,368,373,378,381,418,490,514,519,524,585,597,602,606,610,619,629,643,647,651,656,659,668,680,688,698,711,723,740,754,759,795,798,802,807,871,876,881,885,890,895,899,1040,1056],{"id":42,"type":43,"markdown":44},"intro","prose","You can now measure any angle, draw any angle, and construct 60°, 90°, 30°, 45° and their relatives with a compass and straightedge. You also know *why* those constructions work. This last layer asks: **what can you build with them, and where do they run out?**\n\nYou will construct triangles from just three measurements, draw regular polygons inside a circle, meet a 17-sided shape that a teenager in Germany proved could be constructed, and learn why an innocent-looking angle like **20°** can never be constructed exactly with ruler and compass, no matter how clever you are. Along the way there are projects to make with your hands, olympiad-style puzzles, and a look at the people who use angles for a living.",{"id":46,"type":47,"variant":48,"title":49,"markdown":50},"how-extend","callout","observation","How to use this layer","Pick what interests you: the chapters stand alone. Keep your geometry box handy for the projects. Every number in the worked examples was calculated, so you can check your own drawings against them with a protractor and ruler.",{"id":52,"type":53,"title":54,"eyebrow":55,"navLabel":56},"ch1","chapter","Constructing triangles from three facts","Chapter 01","1 Building triangles",{"id":58,"type":43,"markdown":59},"tri-intro","How much do you need to know about a triangle to draw it exactly? A triangle has six measurements: three sides and three angles. Surprisingly, **three well-chosen facts are enough** to fix the whole triangle. Anyone, anywhere, following the same three facts will draw a copy of exactly the same size and shape.\n\nThe four standard sets of facts are:\n\n- **SSS**: all three sides.\n- **SAS**: two sides and the angle **between** them (the included angle).\n- **ASA**: two angles and the side **between** them.\n- **RHS**: a **right** angle, the **hypotenuse** (the side opposite the right angle) and one other side.\n\nThese are exactly the conditions that make two triangles **congruent** (identical in shape and size), which is the reason the constructions in Deepen work.",{"id":61,"type":62,"title":63,"items":64},"steps-sss","steps","SSS: construct a triangle with sides 5 cm, 6 cm and 7 cm",[65,69,73,77,81],{"title":66,"tag":67,"text":68},"Longest side as base","7 cm","Draw a segment AB = 7 cm with a ruler.",{"title":70,"tag":71,"text":72},"First arc","centre A, 5 cm","Open the compass to 5 cm. With centre A, draw an arc above AB.",{"title":74,"tag":75,"text":76},"Second arc","centre B, 6 cm","Open the compass to 6 cm. With centre B, draw an arc cutting the first arc at C.",{"title":78,"tag":79,"text":80},"Join","AC and BC","Join AC and BC. Triangle ABC has AC = 5 cm, BC = 6 cm, AB = 7 cm.",{"title":82,"tag":83,"text":84},"Check","protractor","Measure the angles. They should be close to 57.1° at A, 44.4° at B and 78.5° at C.",{"id":86,"type":87,"title":88,"problem":89,"steps":90,"help":97},"we-sss-angles","worked_example","What angles should an SSS triangle have?","In the triangle with AB = 7 cm, BC = 6 cm and CA = 5 cm, what should a protractor show at each corner? (This uses the **cosine rule**, a Class 10 idea, just to give you target values.)",[91,92,93,94,95,96],"The cosine rule says: for the angle opposite side c, cos C = (a² + b² − c²) ÷ (2ab).","Angle C (opposite AB = 7): cos C = (5² + 6² − 7²) ÷ (2 × 5 × 6) = 12 ÷ 60 = 0.2, so C ≈ 78.5°.","Angle A (opposite BC = 6): cos A = (5² + 7² − 6²) ÷ (2 × 5 × 7) = 38 ÷ 70, so A ≈ 57.1°.","Angle B (opposite CA = 5): cos B = (6² + 7² − 5²) ÷ (2 × 6 × 7) = 60 ÷ 84, so B ≈ 44.4°.","Check: 57.1 + 44.4 + 78.5 = 180.0°. ✓ The largest angle is opposite the longest side, as always.","If your protractor readings are within 1° of these, your construction is excellent. (You can get the same three numbers without any formula: construct the triangle carefully and measure.)",{"simplerExplanation":98},"You do not need the cosine rule to construct the triangle, only to predict the angles so you can check your drawing.",{"id":100,"type":87,"title":101,"problem":102,"steps":103},"we-sas","SAS: sides 6 cm and 4 cm with 60° between them","Construct triangle PQR with PQ = 6 cm, ∠P = 60° and PR = 4 cm, using only ruler and compass.",[104,105,106,107,108],"Draw PQ = 6 cm.","Construct 60° at P: arc with centre P cuts PQ at X; same radius from X cuts the arc at Y; draw ray PY.","Open the compass to 4 cm, centre P, and cut ray PY at R.","Join QR. Triangle PQR is fixed.","Check by measuring your drawing: QR ≈ 5.29 cm, ∠Q ≈ 40.9°, ∠R ≈ 79.1°. (60 + 40.9 + 79.1 = 180.) These come from the cosine and sine rules, but a careful scale drawing gives the same values.",{"id":110,"type":87,"title":111,"problem":112,"steps":113},"we-asa","ASA: a 7 cm base with 45° and 60° at its ends","Construct triangle LMN with LM = 7 cm, ∠L = 45° and ∠M = 60°.",[114,115,116,117,118],"Draw LM = 7 cm.","At L, construct 45° (build 90°, then bisect it). At M, construct 60° on the same side of LM.","Extend the two rays until they cross. The crossing point is N.","The third angle is ∠N = 180° − 45° − 60° = **75°**, so you can check it with a protractor.","The other sides come out as MN ≈ 5.12 cm and LN ≈ 6.28 cm — measure them on your drawing and you should agree to about a millimetre. The longest side (LM = 7 cm) faces the largest angle (75°).",{"id":120,"type":47,"variant":121,"title":122,"markdown":123},"careful-inequality","careful","Not every three lengths make a triangle","Try SSS with 3 cm, 4 cm and 8 cm. Draw the 8 cm base, then arcs of 3 cm and 4 cm from its ends: they **never meet**, because 3 + 4 = 7 is less than 8. This is the **triangle inequality**: any two sides together must be **longer** than the third. Check this before you start, or you will draw arcs for ever.",{"id":125,"type":47,"variant":126,"title":127,"markdown":128},"misc-ssa","misconception","“Any three facts will do”","Two sides and an angle that is **not** between them (SSA) can give **two different** triangles. For example, with AB = 6 cm, ∠A = 30° and BC = 4 cm, the 4 cm arc from B crosses the other arm of the angle at **two** points, so two different triangles fit the facts. And three angles (AAA) fix the shape but not the size: a small and a large equilateral triangle both have 60°, 60°, 60°.",{"id":130,"type":131,"component":132,"componentVersion":5,"config":133,"objective":158,"textAlternative":159,"help":160},"lab-criteria","interactive","match-pairs",{"prompt":134,"mode":135,"pairs":136},"Match each set of given facts to the criterion that fixes the triangle, or to 'no unique triangle'.","connect",[137,140,143,146,149,152,155],{"a":138,"b":139},"Sides 5 cm, 6 cm, 7 cm","SSS",{"a":141,"b":142},"6 cm, 4 cm and the 60° angle between them","SAS",{"a":144,"b":145},"A 7 cm side with 45° and 60° at its ends","ASA",{"a":147,"b":148},"A right angle, hypotenuse 5 cm, one side 3 cm","RHS",{"a":150,"b":151},"Angles 50°, 60°, 70° and nothing else","AAA: shape only, size not fixed",{"a":153,"b":154},"Sides 3 cm, 4 cm, 8 cm","Impossible: 3 + 4 is less than 8",{"a":156,"b":157},"6 cm, 4 cm and a 30° angle not between them","SSA: can give two triangles","Match triangle data to the congruence criterion that fixes it, or spot data that does not give one triangle.","This game has seven pairs to connect.\n\n- Sides 5, 6, 7 cm ↔ **SSS**.\n- 6 cm, 4 cm and the 60° angle between them ↔ **SAS**.\n- A 7 cm side with 45° and 60° at its ends ↔ **ASA**.\n- A right angle, hypotenuse 5 cm and one side 3 cm ↔ **RHS** (the third side must be 4 cm, because 3² + 4² = 5²).\n- Angles 50°, 60°, 70° only ↔ **AAA fixes the shape but not the size**.\n- Sides 3, 4, 8 cm ↔ **impossible**, because 3 + 4 = 7 is less than 8.\n- 6 cm, 4 cm and a 30° angle not between them ↔ **SSA, which can give two triangles**.\n\nThe lesson: the angle in SAS must be between the two sides, and the side in ASA must be between the two angles.",{"hints":161},[162,163],"Where is the angle compared with the two sides? Between them, or not?","Add the two shorter sides. Is the total bigger than the longest side?",{"id":165,"type":166,"itemId":167,"prompt":168,"check":169,"hints":173,"feedback":176},"practice-sss","practice","constructing-angles.ext-sss-largest","A triangle has sides 5 cm, 6 cm and 7 cm. Which angle is largest, and about how many degrees is it, to the nearest degree? (Enter the number of degrees.)",{"kind":170,"answer":171,"tolerance":5,"unit":172},"number",78,"°",[174,175],"The largest angle is opposite the longest side.","cos C = (25 + 36 − 49) ÷ 60 = 0.2.",{"correct":177,"incorrect":178},"Right: the angle opposite the 7 cm side is about 78.5°.","The largest angle faces the longest side (7 cm). cos C = (5² + 6² − 7²) ÷ (2 × 5 × 6) = 0.2, so C ≈ **78.5°**. Or construct it and measure.",{"id":180,"type":53,"title":181,"eyebrow":182,"navLabel":183},"ch2","Regular polygons from a circle","Chapter 02","2 Regular polygons",{"id":185,"type":43,"markdown":186},"poly-intro","A **regular polygon** has all sides equal and all angles equal. The neat way to draw one is to start with a circle and split the full turn at its centre into equal **central angles**: a regular n-sided polygon has central angle **360° ÷ n**. Mark those points on the circle and join them.\n\nSo drawing a regular polygon with ruler and compass is really the question: **can I construct the central angle 360° ÷ n?**",{"id":188,"type":189,"caption":190,"columns":191,"rows":196},"table-polygons","table","Regular polygons: central angles, interior angles and how to construct them",[192,193,194,195],"Polygon","Central angle","Interior angle","Ruler-and-compass method",[197,202,206,211,214,219,224,229,234],[198,199,200,201],"Triangle (3)","120°","60°","Hexagon points, using every other one",[203,204,204,205],"Square (4)","90°","Two perpendicular diameters",[207,208,209,210],"Pentagon (5)","72°","108°","Possible; Euclid's Elements, Book IV, gives a method",[212,200,199,213],"Hexagon (6)","Step the radius six times round the circle",[215,216,217,218],"Heptagon (7)","51.43°","128.57°","Impossible with ruler and compass",[220,221,222,223],"Octagon (8)","45°","135°","Bisect the square's 90° central angles",[225,226,227,228],"Decagon (10)","36°","144°","Bisect the pentagon's 72° central angles",[230,231,232,233],"Dodecagon (12)","30°","150°","Bisect the hexagon's 60° central angles",[235,236,237,238],"17-gon","21.18°","158.82°","Possible! Gauss, 1796",{"id":240,"type":241,"items":242},"formulas-poly","formulas",[243,246,249,252],{"expression":244,"caption":245},"central angle = 360° ÷ n","The turn at the centre between neighbouring corners.",{"expression":247,"caption":248},"interior angle = (n − 2) × 180° ÷ n","The polygon splits into n − 2 triangles of 180° each.",{"expression":250,"caption":251},"interior + central = 180°","For a regular polygon, e.g. hexagon 120° + 60°.",{"expression":253,"caption":254},"mitre cut = 180° ÷ n","Half the central angle: the cut at each end of a frame piece.",{"id":256,"type":257,"component":258,"componentVersion":5,"config":259,"textAlternative":261},"anim-hexagon","animation","compass-construction",{"construction":260},"angle-60","This animation shows the 60° construction, which is also the heart of the regular hexagon.\n\n1. Draw a ray OA. With centre O and any radius r, draw an arc cutting OA at P.\n2. With the same radius r and centre P, cut the arc at Q. Triangle OPQ is equilateral, so ∠POQ = 60°.\n\n**From 60° to a hexagon:** draw the full circle with centre O and radius r. Starting at P, keep stepping the same radius round the circle: P, Q, then four more points. Because each step turns 60° about the centre, six steps make 6 × 60° = 360° and you land exactly back on P. Join the six points in order and you have a **regular hexagon** whose side equals the radius. Its interior angles are each 120° (two equilateral-triangle angles, 60° + 60°).\n\n**Bonus shapes:** join every other point to get an **equilateral triangle**; bisect each 60° central angle to get 12 points and a **regular dodecagon** with interior angles of 150°.",{"id":263,"type":47,"variant":264,"title":265,"markdown":266},"ex-pentagon","example","The pentagon: the door to 72° and 36°","A regular pentagon has central angles of 360° ÷ 5 = **72°** and interior angles of **108°**. It *can* be constructed with ruler and compass: Euclid gave a method in Book IV of the Elements (Proposition 11), built on the golden ratio. It takes more steps than the hexagon, but it is exact. Once you have 72°, bisecting gives **36°**, and 72° − 60° gives **12°**, a key to the next chapters.",{"id":268,"type":166,"itemId":269,"prompt":270,"check":271,"hints":274,"feedback":277},"practice-octagon","constructing-angles.ext-octagon-interior","What is each interior angle of a regular octagon?",{"kind":170,"answer":272,"tolerance":273,"unit":172},135,0,[275,276],"Use (n − 2) × 180 ÷ n with n = 8.","Or: 180° minus the central angle 45°.",{"correct":278,"incorrect":279},"Right: (8 − 2) × 180 ÷ 8 = 1080 ÷ 8 = 135°.","Interior angle = (8 − 2) × 180° ÷ 8 = 1,080° ÷ 8 = **135°**. Check: 135° + 45° (central angle) = 180°.",{"id":281,"type":43,"markdown":282},"gauss","Which regular polygons *can* be constructed with ruler and compass? The Greeks could do 3, 4, 5, 6, 8, 10, 12, 15 and so on (doubling the sides is easy: just bisect). For two thousand years nobody found another. Then in **1796**, an 18-year-old **Carl Friedrich Gauss** proved that the regular **17-gon** is constructible. The story goes that he was so proud he asked for a 17-gon on his headstone. It never happened: the stonemason said nobody would be able to tell it from a circle, and the monument in Gauss's home town of Brunswick carries a 17-pointed star instead. (It is a much-repeated anecdote rather than a documented request, so treat it as a story.)\n\nThe full answer, completed by **Pierre Wantzel** in 1837, is: a regular n-gon is constructible exactly when n is a **power of 2 times distinct Fermat primes**. The known Fermat primes are **3, 5, 17, 257 and 65,537**. So from 3 to 20 the constructible ones are **3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20**, and the impossible ones are **7, 9, 11, 13, 14, 18, 19**. The heptagon (7) and the nonagon (9 = 3 × 3, a repeated Fermat prime) cannot be drawn exactly.",{"id":284,"type":285,"prompt":286,"options":287,"explanation":300},"predict-9gon","prediction","A regular **nonagon** (9 sides) has a central angle of 360° ÷ 9 = 40°. Can you construct it with ruler and compass?",[288,291,294,297],{"id":289,"label":290},"a","Yes: 40° is a whole number of degrees",{"id":292,"label":293},"b","Yes: 9 = 3 × 3 and 3 is a Fermat prime",{"id":295,"label":296},"c","No: 9 uses the Fermat prime 3 twice, and 40° is not a multiple of 3°",{"id":298,"label":299},"d","Only with a protractor","**No.** Gauss and Wantzel's rule needs *distinct* Fermat primes, and 9 = 3 × 3 repeats one. You can check another way: 40° is not a multiple of 3°, and (as the next chapters show) only whole-degree angles that are multiples of 3° are constructible. Constructing 40° would let you bisect to 20°, which is exactly the famous impossible angle. A protractor can *draw* a very good nonagon, but not an exact construction.",{"id":302,"type":303,"conceptId":304,"relation":305,"explanation":306},"conn-prime","connection","prime-and-composite","related_to","Which regular polygons can be constructed depends on Fermat primes (3, 5, 17, 257, 65,537): a surprising link between prime numbers and ruler-and-compass geometry.",{"id":308,"type":53,"title":309,"eyebrow":310,"navLabel":311},"ch3","The angle that cannot be made: trisection","Chapter 03","3 The trisection story",{"id":313,"type":43,"markdown":314},"trisect-intro","Bisecting any angle is easy. So the ancient Greeks naturally asked: can you **trisect** any angle, cutting it into three equal parts, with only straightedge and compass?\n\nFor some angles, yes. A 90° angle trisects into 30° pieces, and we can construct 30°. A 180° angle trisects into 60° pieces. But the Greeks could never find a method that works for **every** angle, and in particular nobody could trisect **60°** into three 20° pieces.\n\nFor over 2,000 years mathematicians kept trying. In **1837**, the French mathematician **Pierre Wantzel** proved that it is **impossible**. Not just hard: impossible. No sequence of straightedge lines and compass circles, however long, can produce an exact 20° angle from nothing.",{"id":316,"type":47,"variant":317,"title":318,"markdown":319},"aha-why-impossible","aha","Why impossible, in one idea","Every point you can construct is found where lines and circles cross. Working out those crossings only ever needs adding, subtracting, multiplying, dividing and **square roots**. Making a 20° angle turns out to need the solution of a **cubic equation** (one with an x³ in it) that cannot be solved using square roots alone. Circles and lines speak the language of square roots; 20° needs cube-root language. So the tools simply cannot say it.",{"id":321,"type":322,"title":323,"items":324},"timeline-trisection","timeline","Two thousand years of an unsolved puzzle",[325,329,333,337,341,345,349],{"time":326,"title":327,"text":328},"400s BCE","The three problems","Greek geometers are already working on trisecting an angle, doubling a cube and squaring a circle with straightedge and compass. By about 414 BCE “squaring the circle” was familiar enough to be a joke in an Athenian comedy.",{"time":330,"title":331,"text":332},"~300 BCE","Euclid's Elements","Euclid collects constructions, including bisecting angles and the regular pentagon, but no trisection.",{"time":334,"title":335,"text":336},"~250 BCE","Archimedes cheats cleverly","Archimedes trisects any angle using a ruler with two marks on it (a neusis construction), breaking the rules on purpose.",{"time":338,"title":339,"text":340},"1796","Gauss's 17-gon","Gauss constructs the regular 17-gon and links constructibility to Fermat primes.",{"time":342,"title":343,"text":344},"1837","Wantzel's proof","Wantzel proves general trisection and doubling the cube are impossible with straightedge and compass.",{"time":346,"title":347,"text":348},"1882","Squaring the circle","Lindemann proves π is transcendental, so squaring the circle is impossible too.",{"time":350,"title":351,"text":352},"1980","Origami trisection","A paper-folding trisection is reported, due to Hisashi Abe: folds can do what circles and lines cannot.",{"id":354,"type":87,"title":355,"problem":356,"steps":357},"we-neusis","Archimedes' marked-ruler trisection (for a 60° angle)","Archimedes allowed himself one extra move: a ruler with **two marks** on it, a distance r apart, that can be slid into position. Here is his method, for ∠AOB = 60°.",[358,359,360,361,362,363],"Draw a circle with centre O and radius r, the same as the distance between the ruler's two marks. Let B be where arm OB meets the circle.","Extend arm OA backwards through O as a straight line.","Slide the marked ruler so that it passes through B, one mark lies on the circle (call it C) and the other mark lies on the extended line (call it D). So CD = r.","Then ∠CDO is exactly **one third** of ∠AOB: here 60° ÷ 3 = **20°**.","Why: triangles DCO and COB are isosceles (DC = CO = r and CO = OB = r). Chasing the angles shows ∠AOB = 3 × ∠CDO.","The catch: sliding a marked ruler until two conditions hold is not allowed in Greek constructions. The trick is exact, but it breaks the rules.",{"id":365,"type":47,"variant":126,"title":366,"markdown":367},"misc-trisect-chord","“Just split the chord into three”","A tempting method: join the ends of the arc (the chord), divide that straight chord into three equal parts, and join the division points to the vertex. For 60° this gives pieces of about 19.11°, 21.79° and 19.11°, not 20°, 20°, 20°. The middle piece is too big, because the chord is closer to the vertex in the middle than at the ends. It is a decent **approximation** — the two outer pieces are about 0.9° too small and the middle one about 1.8° too big — but mathematicians call it wrong, because *exact* is the whole point. Every year people send \"trisection proofs\" to universities; almost all are approximations like this one.",{"id":369,"type":47,"variant":370,"title":371,"markdown":372},"try-origami","try_it","Folding beats the compass","Paper folding follows different rules. One allowed fold places two given points onto two given lines at the same time, and that single fold can solve cubic equations. So **origami can trisect any angle** and can even fold a regular heptagon. Look up \"Abe's origami trisection\" and try it with a square of paper and a 60° angle: you will fold an exact 20°.",{"id":374,"type":53,"title":375,"eyebrow":376,"navLabel":377},"ch4","Which whole-degree angles can you construct?","Chapter 04","4 Multiples of 3°",{"id":379,"type":43,"markdown":380},"mult3","Here is a beautiful, complete answer. Among whole numbers of degrees, **you can construct exactly the multiples of 3°**, and nothing else.\n\n**Why every multiple of 3° is possible.** You can construct 60° (equilateral triangle) and 72° (regular pentagon). Subtract: 72° − 60° = **12°**. Bisect: **6°**. Bisect again: **3°**. Once you have 3°, you can copy it side by side as many times as you like: 6°, 9°, 12°, … every multiple of 3°.\n\n**Why nothing else is possible.** If you could construct any whole-degree angle that is *not* a multiple of 3°, you could combine it with 3° (adding and subtracting copies) to get **1°**. Then twenty copies of 1° would give **20°**. But Wantzel proved 20° impossible. So 1°, 2°, 4°, 5°, 10°, 20°, 40°, 50°, 70°, 100°… are all out of reach.",{"id":382,"type":189,"caption":383,"columns":384,"rows":387},"table-mult3","A route to some surprising constructible angles",[385,386,82],"Angle","Route",[388,391,394,398,402,406,410,414],[208,389,390],"Central angle of the regular pentagon","360 ÷ 5 = 72",[226,392,393],"Bisect 72°","72 ÷ 2 = 36",[395,396,397],"12°","72° take away 60°","72 − 60 = 12",[399,400,401],"18°","Bisect 36°","36 ÷ 2 = 18",[403,404,405],"3°","18° − 15° (or bisect 12° twice)","18 − 15 = 3",[407,408,409],"9°","Three copies of 3°, or 45° − 36°","45 − 36 = 9",[411,412,413],"81°","90° − 9°","90 − 9 = 81",[415,416,417],"54°","90° − 36°","90 − 36 = 54",{"id":419,"type":131,"component":420,"componentVersion":5,"config":421,"objective":483,"textAlternative":484,"help":485},"lab-constructible","sort-game",{"prompt":422,"bins":423,"items":430,"seconds":273},"Can this whole-degree angle be constructed exactly with straightedge and compass?",[424,427],{"id":425,"label":426},"yes","Constructible",{"id":428,"label":429},"no","Not constructible",[431,434,437,440,443,446,449,452,455,459,463,467,471,475,479],{"id":432,"label":403,"bin":425,"why":433},"a3","72° − 60° = 12°, then bisect twice: 12 → 6 → 3.",{"id":435,"label":407,"bin":425,"why":436},"a9","45° − 36°, or three copies of 3°. 9 is a multiple of 3.",{"id":438,"label":395,"bin":425,"why":439},"a12","The pentagon's 72° minus the equilateral 60°.",{"id":441,"label":399,"bin":425,"why":442},"a18","Bisect 36°, which is half of the pentagon's 72°.",{"id":444,"label":226,"bin":425,"why":445},"a36","Half of 72°; also the tip angle of the star pentagon.",{"id":447,"label":415,"bin":425,"why":448},"a54","90° − 36°. 54 = 3 × 18.",{"id":450,"label":208,"bin":425,"why":451},"a72","The central angle of a regular pentagon, which Euclid constructed.",{"id":453,"label":411,"bin":425,"why":454},"a81","90° − 9°. 81 = 3 × 27, a multiple of 3.",{"id":456,"label":457,"bin":428,"why":458},"a1","1°","Not a multiple of 3. Twenty copies would give 20°, which is impossible.",{"id":460,"label":461,"bin":428,"why":462},"a10","10°","Two copies would give 20°, the impossible trisection of 60°.",{"id":464,"label":465,"bin":428,"why":466},"a20","20°","One third of 60°: Wantzel proved this cannot be constructed (1837).",{"id":468,"label":469,"bin":428,"why":470},"a40","40°","Bisecting 40° would give 20°. It is the nonagon's central angle, also impossible.",{"id":472,"label":473,"bin":428,"why":474},"a50","50°","50 is not a multiple of 3, so combined with 3° it would lead to 1° and then 20°.",{"id":476,"label":477,"bin":428,"why":478},"a70","70°","Not a multiple of 3. 70 = 3 × 23 + 1, so it would give 1°.",{"id":480,"label":481,"bin":428,"why":482},"a100","100°","Not a multiple of 3. 100 − 99 = 1°, and 99° is constructible, so 1° would follow.","Sort whole-degree angles into those you can construct exactly with straightedge and compass and those you cannot.","This sort game has 15 angle cards and two bins.\n\n**Constructible:** 3°, 9°, 12°, 18°, 36°, 54°, 72° and 81°. Every one is a multiple of 3°. Routes: 72° is the pentagon's central angle; 72° − 60° = 12°; bisecting 12° twice gives 3°; half of 72° is 36°, and half of 36° is 18°; 90° − 36° = 54°; 45° − 36° = 9°; 90° − 9° = 81°.\n\n**Not constructible:** 1°, 10°, 20°, 40°, 50°, 70° and 100°. None is a multiple of 3°. Any of them, combined with the constructible 3°, would produce 1°, and 20 copies of 1° would make 20°, which Wantzel proved impossible in 1837.\n\nThe quick test: divide by 3. A whole number means constructible.",{"simplerExplanation":486,"hints":487},"Divide the angle by 3. If it goes exactly, you can construct it. If not, you cannot.",[488,489],"Add up the digits: if the digit sum is a multiple of 3, so is the number.","20° is the famous impossible angle.",{"id":491,"type":166,"itemId":492,"prompt":493,"check":494,"hints":509,"feedback":511},"practice-mult3","constructing-angles.ext-which-constructible","Which of these angles **can** be constructed exactly with straightedge and compass? Choose all that apply.",{"kind":495,"options":496,"correct":508},"choice",[497,499,501,503,505],{"id":289,"label":498},"27°",{"id":292,"label":500},"35°",{"id":295,"label":502},"48°",{"id":298,"label":504},"80°",{"id":506,"label":507},"e","105°",[289,295,506],[510],"Test each number: is it divisible by 3?",{"correct":512,"incorrect":513},"Right: 27, 48 and 105 are multiples of 3; 35 and 80 are not.","Divide by 3: 27 ÷ 3 = 9 ✓, 48 ÷ 3 = 16 ✓, 105 ÷ 3 = 35 ✓. But 35 and 80 leave remainders, so they cannot be constructed exactly. Answer: **27°, 48° and 105°**.",{"id":515,"type":47,"variant":516,"title":517,"markdown":518},"nuance-draw-vs-construct","nuance","Impossible to construct, easy to draw","\"20° cannot be constructed\" does **not** mean you cannot draw 20°! A protractor draws it in seconds, to within a degree. It means there is no **exact** straightedge-and-compass recipe. In real life, measuring tools are fine. The impossibility is about what pure reasoning with circles and lines can reach.",{"id":520,"type":53,"title":521,"eyebrow":522,"navLabel":523},"ch5","Angles at work in the real world","Chapter 05","5 Real-world angles",{"id":525,"type":526,"title":527,"prompt":528,"options":529},"explorer-real","explorer","Where construction angles earn a living","Pick a context to see which angles matter and how they are made.",[530,541,552,563,574],{"id":531,"label":532,"chain":533,"badge":538,"note":540},"carpentry","Carpentry",[534,535,536,537],"Plan a frame","Mitre = 180° ÷ n","Set the saw","Pieces meet exactly",{"text":539,"tone":425},"Uses constructed angles","A picture frame has 4 sides, so each end is cut at 180° ÷ 4 = **45°**; two 45° cuts meet in a 90° corner. A hexagonal planter needs 180° ÷ 6 = **30°** cuts, and an octagonal gazebo 180° ÷ 8 = **22.5°**. Carpenters use a mitre box or mitre saw with these angles marked, and a try square to check right angles.",{"id":542,"label":543,"chain":544,"badge":549,"note":551},"architecture","Architecture",[545,546,547,548],"Brief","Scale drawing","Roof pitch, stairs","Building",{"text":550,"tone":425},"Angles on every drawing","Roofs are pitched so monsoon rain runs off fast; staircases have a comfortable, safe angle; ramps for wheelchairs must be gentle. Architects once drew all of this with set squares and compasses on drawing boards; today CAD software constructs the same lines exactly.",{"id":553,"label":554,"chain":555,"badge":560,"note":562},"roads","Road design",[556,557,558,559],"Traffic study","Junction layout","Sight lines","Safe crossing",{"text":561,"tone":425},"Angles save lives","Road designers try to make roads meet at close to **90°**, because drivers at a sharp, skewed junction cannot see traffic coming. Hill roads use hairpin bends to keep slopes gentle, and railway points meet at very small angles so wheels can switch tracks smoothly.",{"id":564,"label":565,"chain":566,"badge":571,"note":573},"sports","Sports fields",[567,568,569,570],"Mark a baseline","3-4-5 rope","Right-angle corner","Paint the lines",{"text":572,"tone":425},"Right angles with a rope","Groundskeepers lay out a kabaddi court or a football pitch with a rope knotted into **12 equal parts**. Pegged out as a triangle with sides of 3, 4 and 5 parts, it makes a perfect right angle, because 3² + 4² = 9 + 16 = 25 = 5². The Sulba Sutras of ancient India used ropes and pegs in just this spirit to lay out altars.",{"id":575,"label":576,"chain":577,"badge":582,"note":584},"kites","Kite making",[578,579,580,581],"Spine stick","Bent bow","Tie at 90°","Balance",{"text":583,"tone":425},"Perpendicular bisector","A patang's straight spine crosses the curved bow at **right angles**, and the spine must hit the bow's midpoint: that is a perpendicular bisector. A lopsided frame catches unequal wind and spins. Kite fighters tune the bridle string angle to control how the kite dives and climbs.",{"id":586,"type":166,"itemId":587,"prompt":588,"check":589,"hints":591,"feedback":594},"practice-mitre","constructing-angles.ext-mitre-hexagon","A carpenter is making a frame shaped like a **regular hexagon**. At what angle should each end of each piece be cut (the mitre angle)?",{"kind":170,"answer":590,"tolerance":273,"unit":172},30,[592,593],"Mitre angle = 180° ÷ n.","A picture frame (n = 4) uses 45° cuts.",{"correct":595,"incorrect":596},"Right: 180° ÷ 6 = 30°.","For a regular n-sided frame the mitre angle is 180° ÷ n, so 180° ÷ 6 = **30°**. (Each cut is 30° off square; the two pieces then meet at the hexagon's 120° corner.)",{"id":598,"type":53,"title":599,"eyebrow":600,"navLabel":601},"ch6","Projects to make","Chapter 06","6 Projects",{"id":603,"type":47,"variant":370,"title":604,"markdown":605},"try-paper-protractor","Project 1: a paper protractor by folding","1. Cut a circle from paper using a compass (or trace a steel plate). Fold it in half: the crease is a diameter, and each half is 180°.\n2. Fold the semicircle in half: the new crease makes **90°**.\n3. Fold again: **45°** marks. Once more: **22.5°** marks.\n4. Label the creases 0°, 22.5°, 45°, 67.5°, 90°, … up to 180°.\n\nFolding in half is bisecting, so every crease is exact. Now compare your paper protractor with the real one. Challenge: can you get 60° by folding? (Hint: fold the edge of the circle over so it just touches the centre. The crease meets the circle at two points 120° apart; halve that.)",{"id":607,"type":47,"variant":370,"title":608,"markdown":609},"try-clinometer","Project 2: a clinometer from a protractor","Tape a drinking straw along the base line of your protractor. Tie a thread to the centre point and hang a small weight (an eraser or a nut) on it. Look at the top of a tree or a building through the straw. The thread hangs straight down, and the angle between the thread and the 90° mark tells you the **angle of elevation**. If the thread crosses at 45° on the scale, you are looking up at 90° − 45° = 45°. Work in pairs: one sights, one reads.",{"id":611,"type":87,"title":612,"problem":613,"steps":614},"we-tree-45","How tall is the neem tree?","Meera stands 10 m from a neem tree. Through her clinometer the angle of elevation to the top is **45°**. Her eyes are 1.4 m above the ground. How tall is the tree?",[615,616,617,618],"At 45°, the right triangle from her eye to the treetop has two equal sides: it is a 45°–45°–90° triangle, like half a square.","So the height above eye level equals the distance to the tree: **10 m**.","Add her eye height: 10 m + 1.4 m = **11.4 m**.","Trick: walk backwards or forwards until the clinometer reads exactly 45°, then just pace the distance. No calculation needed.",{"id":620,"type":87,"title":621,"problem":622,"steps":623},"we-tree-30","The same idea at 30°","Standing 20 m from a mobile tower, Arjun measures an angle of elevation of 30°. His eyes are 1.4 m up. How tall is the tower, roughly?",[624,625,626,627,628],"Draw it to scale: a base of 20 m (say 10 cm on paper) and a 30° angle constructed at one end (bisect a 60°).","Draw a vertical line at the other end and measure where it meets the sloping line. At this scale it is about 5.77 cm, which is 11.55 m.","(With trigonometry, a Class 10 idea: height above eye = 20 × tan 30° ≈ 11.55 m.)","Add the eye height: 11.55 + 1.4 ≈ **12.9 m**.","Scale drawings with constructed angles let you measure things you could never reach with a tape.",{"id":630,"type":166,"itemId":631,"prompt":632,"check":633,"hints":637,"feedback":640},"practice-tree","constructing-angles.ext-tree-height","Standing 15 m from a building, you measure an angle of elevation of exactly 45° to the roof. Your eyes are 1.5 m above the ground. How tall is the building, in metres?",{"kind":170,"answer":634,"tolerance":635,"unit":636},16.5,0.05,"m",[638,639],"At 45°, height above eye level = distance from the building.","Then add your eye height.",{"correct":641,"incorrect":642},"Right: 15 + 1.5 = 16.5 m.","At 45° the height above your eyes equals the distance, 15 m. Add 1.5 m: **16.5 m**.",{"id":644,"type":47,"variant":370,"title":645,"markdown":646},"try-rangoli-ext","Project 3: a constructed rangoli","Draw a circle and step the radius round it to get six points (60° apart). Bisect each gap to get twelve points (30° apart). Now join points in patterns: every 2nd point (a hexagon), every 3rd (a square, because 3 steps of 30° is 90°; there are three such squares), every 4th (equilateral triangles), every 5th (a twelve-pointed star). Colour the regions. Which patterns come back to the start after the fewest lines, and why? (Think about how 12 shares factors with the step size.)",{"id":648,"type":47,"variant":121,"title":649,"markdown":650},"careful-projects","Safety for outdoor projects","When measuring trees or buildings, stand on safe, level ground away from roads, and never look at the Sun through a straw or clinometer. Keep compass points capped when carrying your geometry box outdoors.",{"id":652,"type":53,"title":653,"eyebrow":654,"navLabel":655},"ch7","Olympiad-style puzzles","Chapter 07","7 Puzzles",{"id":657,"type":43,"markdown":658},"puzzles-intro","These problems mix angle facts, constructions and careful reasoning. Try each before reading the solution. Olympiad problems reward a clear diagram and patient step-by-step thinking more than clever tricks.",{"id":660,"type":87,"title":661,"problem":662,"steps":663},"we-clock","The clock at 3:40","What is the smaller angle between the hands of a clock at **3:40**? What is the reflex angle? Could you construct the smaller one exactly?",[664,665,666,667],"The minute hand moves 360° ÷ 60 = 6° per minute. At 40 minutes it is at 40 × 6° = 240° from 12.","The hour hand moves 30° per hour **and** 0.5° per minute (30° ÷ 60). At 3:40 it is at 3 × 30° + 40 × 0.5° = 90° + 20° = 110°.","Difference: 240° − 110° = **130°**. The reflex angle is 360° − 130° = **230°**.","Is 130° constructible? 130 ÷ 3 = 43⅓, not a whole number, so **no**: it can be drawn with a protractor but not constructed exactly.",{"id":669,"type":166,"itemId":670,"prompt":671,"check":672,"hints":674,"feedback":677},"practice-clock","constructing-angles.ext-clock-1010","What is the smaller angle between the hands of a clock at **10:10**?",{"kind":170,"answer":673,"tolerance":273,"unit":172},115,[675,676],"Minute hand: 10 × 6° = 60° from 12.","Hour hand: 10 × 30° + 10 × 0.5° = 305° from 12.",{"correct":678,"incorrect":679},"Right: 305° − 60° = 245°, and the smaller angle is 360° − 245° = 115°.","Minute hand at 60°, hour hand at 300° + 5° = 305°. The gap is 245°, so the smaller angle is 360° − 245° = **115°**.",{"id":681,"type":87,"title":682,"problem":683,"steps":684},"we-bisections","How many halvings?","Starting from a constructed 60°, how many bisections do you need to reach 3.75°? Is 3.75° a whole number of degrees, and does that matter?",[685,686,687],"Each bisection halves: 60 → 30 → 15 → 7.5 → 3.75.","That is **4** bisections, because 60 ÷ 2⁴ = 60 ÷ 16 = 3.75.","3.75° is not a whole number, but it is still constructible: bisection always works. The multiples-of-3° rule only sorts **whole-degree** angles.",{"id":689,"type":87,"title":690,"problem":691,"steps":692},"we-75-fewest","75° in as few arcs as possible","Construct 75° at O on ray OA using as few compass arcs as you can.",[693,694,695,696,697],"75° lies halfway between 60° and 90°: (60 + 90) ÷ 2 = 75.","Big arc from O cutting OA at P (arc 1). From P, cut at Q, the 60° mark (arc 2). From Q, cut at R, the 120° mark (arc 3).","Arcs from Q and R meet at S; OS is 90° (arcs 4 and 5). Let OS cross the big arc at U.","Arcs from Q and U with equal radius meet at T (arcs 6 and 7). Ray OT bisects 60°–90°, so ∠AOT = **75°**.","Seven arcs. Challenge: 75° = 180° − 105°; can a different route use fewer? Compare your record with a friend's.",{"id":699,"type":285,"prompt":700,"options":701,"explanation":710},"predict-reflex-puzzle","Two arms make an ordinary angle of 105°. Which is true about the **reflex** angle between them?",[702,704,706,708],{"id":289,"label":703},"It is 255° and it is constructible",{"id":292,"label":705},"It is 255° and it is not constructible",{"id":295,"label":707},"It is 75°",{"id":298,"label":709},"It is 285°","**255°, and constructible.** 360° − 105° = 255°. 105° = 90° + 15° is constructible, and the reflex angle is just the other side of the same two arms, so it comes for free. Check with the rule: 255 ÷ 3 = 85 exactly.",{"id":712,"type":166,"itemId":713,"prompt":714,"check":715,"hints":717,"feedback":720},"practice-bisect-count","constructing-angles.ext-bisect-90","Starting from a constructed 90°, how many bisections give 5.625°?",{"kind":170,"answer":716,"tolerance":273},4,[718,719],"Keep halving: 90 → 45 → …","90 ÷ 2ⁿ = 5.625.",{"correct":721,"incorrect":722},"Right: 90 → 45 → 22.5 → 11.25 → 5.625, four bisections (90 ÷ 16).","Halve repeatedly: 90 → 45 → 22.5 → 11.25 → 5.625. That is **4** bisections, since 90 ÷ 16 = 5.625.",{"id":724,"type":131,"component":83,"componentVersion":5,"config":725,"objective":733,"textAlternative":734,"help":735},"lab-construct-ext",{"mode":726,"targets":727,"tolerance":5},"construct",[728,729,730,731,732],15,165,195,285,345,"Draw five tricky angles, including reflex ones, to within 1° using a virtual protractor.","This lab gives one arm and a target; you set the second arm with a protractor. Within 1° scores.\n\nTargets and smart routes:\n\n- **15°**: very thin, a quarter of 60°.\n- **165°**: almost straight; draw 180° − 15°, i.e. 15° short of a straight line.\n- **195°**: reflex; 180° + 15°. Extend the first arm backwards and turn 15° further.\n- **285°**: reflex; 360° − 75°. Draw 75° and take the outside.\n- **345°**: reflex; 360° − 15°. Only 15° short of a full turn.\n\nEvery one is a multiple of 15°, so every one could also be constructed exactly with ruler and compass.",{"simplerExplanation":736,"hints":737},"For reflex targets, work out 360° minus the target, draw that small angle, and use the outside.",[738,739],"195° = 180° + 15°.","345° = 360° − 15°.",{"id":741,"type":131,"component":742,"componentVersion":5,"config":743,"objective":748,"textAlternative":749,"help":750},"lab-estimate-ext","angle-lab",{"modes":744,"allowReflex":746,"rounds":747},[745],"estimate",true,12,"Estimate twelve angles, including reflex angles, as accurately as you can by eye.","This game shows twelve angles between 0° and 360°, one at a time. You type an estimate; the game reveals the true value and scores by closeness.\n\nExpert strategies: for a reflex angle, estimate the small angle outside it and subtract from 360° (an outside gap of about 40° means about 320°). Use benchmarks at every 45°: 45, 90, 135, 180, 225, 270, 315. Picture a clock: every hour is 30°. Surveyors and pilots train this skill because a good estimate catches instrument mistakes.\n\nTry to get your average error below 10°, then below 5°.",{"hints":751},[752,753],"Estimate the gap to the nearest straight line or right angle, then adjust.","Reflex: 360° minus the outside angle.",{"id":755,"type":53,"title":756,"eyebrow":757,"navLabel":758},"ch8","People who work with angles","Chapter 08","8 Careers",{"id":760,"type":189,"caption":761,"columns":762,"rows":766},"table-careers","Careers where measuring and constructing angles matter",[763,764,765],"Career","How angles appear","Tools today",[767,771,775,779,783,787,791],[768,769,770],"Carpenter \u002F furniture maker","Mitre joints, dovetails, chair leg splay","Mitre saw, sliding bevel, try square",[772,773,774],"Architect","Roof pitch, stairs, ramps, building plans","CAD software, laser measures",[776,777,778],"Civil engineer","Road junctions, bridge trusses, slopes of embankments","CAD, survey data, simulation",[780,781,782],"Surveyor","Measuring land boundaries and heights by angles","Total station, GPS, theodolite",[784,785,786],"Draughtsperson \u002F CAD designer","Exact technical drawings of machines and parts","CAD programs that construct with lines and circles",[788,789,790],"Game and animation designer","Rotating characters, cameras and lighting angles","3D software; angles in code",[792,793,794],"Pilot \u002F sailor \u002F navigator","Headings and bearings measured in degrees from north","Compass, GPS, charts",{"id":796,"type":43,"markdown":797},"cad","A striking fact: modern **CAD** (computer-aided design) programs used by engineers are, underneath, digital versions of straightedge and compass. You pick points, draw lines through them and circles around them, and the program finds where they cross, exactly as Euclid did. A **total station**, the yellow instrument on a tripod you may see beside a new road, is a super-accurate protractor combined with a laser rangefinder. It measures angles to within a few *seconds* of arc (a second is 1\u002F3,600 of a degree).",{"id":799,"type":47,"variant":264,"title":800,"markdown":801},"ex-sulba","Ancient Indian surveyors","The **Sulba Sutras** (the oldest parts are often dated to around 800–500 BCE) were manuals for building Vedic fire altars of exact shapes and areas. Using only a rope (sulba) and pegs, the priests could draw a perpendicular, a square, and shapes of equal area. Their right-angle rules match the triples 3-4-5 and 5-12-13 (25 + 144 = 169), centuries before Greek geometry books.",{"id":803,"type":53,"title":804,"eyebrow":805,"navLabel":806},"ch9","Words and links for the wider picture","Chapter 09","9 Words and links",{"id":808,"type":809,"title":810,"terms":811},"glossary-extend","glossary","Extend vocabulary",[812,815,817,819,821,824,828,831,835,839,842,846,849,853,856,859,862,865,868],{"term":139,"meaning":813,"example":814},"Three sides known: fixes a unique triangle (if the triangle inequality holds).","5 cm, 6 cm, 7 cm",{"term":142,"meaning":816},"Two sides and the included angle (the angle between them): fixes a unique triangle.",{"term":145,"meaning":818},"Two angles and the included side (the side between them): fixes a unique triangle.",{"term":148,"meaning":820},"Right angle, hypotenuse and one other side: fixes a unique right triangle.",{"term":822,"meaning":823},"hypotenuse","The longest side of a right triangle, opposite the right angle.",{"term":825,"meaning":826,"example":827},"triangle inequality","In any triangle, the sum of any two sides is greater than the third side.","3, 4, 8 cm cannot make a triangle.",{"term":829,"meaning":830},"congruent","Exactly the same shape and size, so one would fit perfectly on the other.",{"term":832,"meaning":833,"example":834},"regular polygon","A polygon with all sides equal and all angles equal.","A square, a regular hexagon.",{"term":836,"meaning":837,"example":838},"central angle","The angle at the centre between two neighbouring corners of a regular polygon: 360° ÷ n.","72° for a pentagon.",{"term":840,"meaning":841},"interior angle","An angle inside a polygon at a corner; for a regular n-gon it is (n − 2) × 180° ÷ n.",{"term":843,"meaning":844,"example":845},"constructible angle","An angle that can be made exactly with straightedge and compass alone.","3°, 15°, 72°",{"term":847,"meaning":848},"trisect","To divide into three equal parts.",{"term":850,"meaning":851,"example":852},"neusis","A 'sliding' construction using a ruler with two marks; not allowed in classical constructions.","Archimedes' trisection",{"term":854,"meaning":855},"Fermat prime","A prime of the form 2 raised to a power of 2, plus 1. Known ones: 3, 5, 17, 257, 65,537.",{"term":857,"meaning":858},"origami construction","Making exact shapes and angles by folding paper; it can trisect angles.",{"term":860,"meaning":861},"clinometer","An instrument for measuring angles of slope or elevation.",{"term":863,"meaning":864},"angle of elevation","The angle you look up through, measured from the horizontal.",{"term":866,"meaning":867},"mitre joint","A corner joint where two pieces are cut at equal angles, such as 45° for a picture frame.",{"term":869,"meaning":870},"total station","A surveying instrument that measures angles and distances very precisely.",{"id":872,"type":303,"conceptId":873,"relation":874,"explanation":875},"conn-shape-ext","shape-and-space","applied_in","Constructing triangles and regular polygons turns angle skills into exact shapes: hexagons, octagons, and triangles fixed by SSS, SAS or ASA.",{"id":877,"type":303,"conceptId":878,"relation":879,"explanation":880},"conn-angles-ext","angles","helps_understand","Angle sums (180° in a triangle, 360° round a point) power every calculation here, from ASA third angles to clock-hand puzzles.",{"id":882,"type":303,"conceptId":883,"relation":305,"explanation":884},"conn-patterns-ext","patterns","Halving angles (60, 30, 15, 7.5…) and stepping round a circle in rangoli designs are number and shape patterns.",{"id":886,"type":53,"title":887,"eyebrow":888,"navLabel":889},"ch10","Open questions and wrap-up","Chapter 10","10 Wrap-up",{"id":891,"type":47,"variant":892,"title":893,"markdown":894},"open-questions","question","Open questions to explore","Some of these have known answers you can research; some are still open.\n\n- Are there more **Fermat primes** after 65,537? Nobody knows. If one exists, another regular polygon becomes constructible.\n- What can you construct with a **compass alone** (no straightedge)? Mohr and Mascheroni proved: everything a ruler-and-compass can. Why is that believable?\n- What can you do with a **straightedge alone**? Much less. What is missing?\n- Paper folding can trisect angles. Which regular polygons can origami make that compasses cannot?\n- How accurate must a surveyor's angle be to build a 100 m bridge that meets in the middle from both banks?\n- Why do honeycombs use 120° angles and not 90°? Is it about wax, strength or both?",{"id":896,"type":897,"prompt":898},"reflect-extend","reflection","Wantzel proved that 20° can never be constructed. Explain in your own words why a *proof* of impossibility is more powerful than 2,000 years of failed attempts. Then describe one situation in real life where \"impossible to construct exactly\" does not matter at all, and one where exactness really matters.",{"id":900,"type":901,"title":9,"questions":902},"quiz-extend","quiz",[903,916,929,941,950,960,972,985,998,1009,1020,1031],{"itemId":904,"prompt":905,"options":906,"correct":289,"why":915},"constructing-angles.ext-q-sas","Which set of facts fixes a unique triangle?",[907,909,911,913],{"id":289,"label":908},"Two sides and the angle between them",{"id":292,"label":910},"Three angles",{"id":295,"label":912},"Two sides and an angle not between them",{"id":298,"label":914},"One side and one angle","That is SAS. AAA fixes only the shape; SSA can give two triangles.",{"itemId":917,"prompt":918,"options":919,"correct":292,"why":928},"constructing-angles.ext-q-inequality","Which three lengths can form a triangle?",[920,922,924,926],{"id":289,"label":921},"2 cm, 3 cm, 6 cm",{"id":292,"label":923},"4 cm, 5 cm, 8 cm",{"id":295,"label":925},"1 cm, 2 cm, 3 cm",{"id":298,"label":927},"3 cm, 4 cm, 8 cm","4 + 5 = 9 is greater than 8. In (c), 1 + 2 = 3 exactly, so the triangle collapses flat.",{"itemId":930,"prompt":931,"options":932,"correct":292,"why":940},"constructing-angles.ext-q-asa","A triangle has angles 45° and 60° at the ends of a 7 cm side. What is the third angle?",[933,935,937,939],{"id":289,"label":934},"65°",{"id":292,"label":936},"75°",{"id":295,"label":938},"85°",{"id":298,"label":507},"180° − 45° − 60° = 75°.",{"itemId":942,"prompt":943,"options":944,"correct":292,"why":949},"constructing-angles.ext-q-central","What is the central angle of a regular pentagon?",[945,946,947,948],{"id":289,"label":200},{"id":292,"label":208},{"id":295,"label":209},{"id":298,"label":226},"360° ÷ 5 = 72°. The interior angle is 108°.",{"itemId":951,"prompt":952,"options":953,"correct":295,"why":959},"constructing-angles.ext-q-dodecagon","What is each interior angle of a regular 12-gon?",[954,955,956,957],{"id":289,"label":199},{"id":292,"label":222},{"id":295,"label":232},{"id":298,"label":958},"160°","(12 − 2) × 180° ÷ 12 = 1,800° ÷ 12 = 150°.",{"itemId":961,"prompt":962,"options":963,"correct":295,"why":971},"constructing-angles.ext-q-heptagon","Which regular polygon can NOT be constructed with ruler and compass?",[964,966,968,970],{"id":289,"label":965},"Pentagon",{"id":292,"label":967},"Octagon",{"id":295,"label":969},"Heptagon",{"id":298,"label":235},"7 is not a Fermat prime and not a power of 2, so the heptagon is impossible. Gauss showed the 17-gon is possible.",{"itemId":973,"prompt":974,"options":975,"correct":295,"why":984},"constructing-angles.ext-q-gauss","Who proved the regular 17-gon is constructible, and when?",[976,978,980,982],{"id":289,"label":977},"Euclid, about 300 BCE",{"id":292,"label":979},"Archimedes, about 250 BCE",{"id":295,"label":981},"Gauss, 1796",{"id":298,"label":983},"Wantzel, 1837","Gauss was 18 at the time. Wantzel later completed the proof of which polygons are impossible.",{"itemId":986,"prompt":987,"options":988,"correct":292,"why":997},"constructing-angles.ext-q-20","Why can 20° not be constructed?",[989,991,993,995],{"id":289,"label":990},"It is too small to draw",{"id":292,"label":992},"It would trisect 60°, which requires solving a cubic that square roots cannot",{"id":295,"label":994},"Protractors do not show it",{"id":298,"label":996},"Nobody has tried hard enough","Wantzel (1837) proved trisecting 60° is impossible with straightedge and compass.",{"itemId":999,"prompt":1000,"options":1001,"correct":295,"why":1008},"constructing-angles.ext-q-mult3","Which whole-degree angle is constructible?",[1002,1004,1005,1007],{"id":289,"label":1003},"25°",{"id":292,"label":469},{"id":295,"label":1006},"57°",{"id":298,"label":477},"57 = 3 × 19, a multiple of 3. The others are not multiples of 3.",{"itemId":1010,"prompt":1011,"options":1012,"correct":295,"why":1019},"constructing-angles.ext-q-mitre","A regular octagonal frame needs mitre cuts of:",[1013,1014,1015,1017],{"id":289,"label":221},{"id":292,"label":231},{"id":295,"label":1016},"22.5°",{"id":298,"label":1018},"67.5°","Mitre = 180° ÷ n = 180° ÷ 8 = 22.5°.",{"itemId":1021,"prompt":1022,"options":1023,"correct":292,"why":1030},"constructing-angles.ext-q-clock","What is the smaller angle between clock hands at 3:40?",[1024,1025,1027,1029],{"id":289,"label":199},{"id":292,"label":1026},"130°",{"id":295,"label":1028},"140°",{"id":298,"label":232},"Minute hand at 240°, hour hand at 90° + 20° = 110°; 240 − 110 = 130°.",{"itemId":1032,"prompt":1033,"options":1034,"correct":292,"why":1039},"constructing-angles.ext-q-rope","A rope knotted into 12 equal parts is pegged as a triangle with sides 3, 4 and 5 parts. What angle is opposite the 5-part side?",[1035,1036,1037,1038],{"id":289,"label":200},{"id":292,"label":204},{"id":295,"label":199},{"id":298,"label":221},"3² + 4² = 25 = 5², so the triangle is right-angled, with the right angle opposite the longest side.",{"id":1041,"type":1042,"title":1043,"points":1044},"cheat-extend","summary","Cheat sheet",[1045,1046,1047,1048,1049,1050,1051,1052,1053,1054,1055],"**SSS, SAS, ASA, RHS** each fix a unique triangle. The angle in SAS and the side in ASA must be **between** the other two facts.","**Triangle inequality:** any two sides must add to more than the third (3, 4, 8 fails). **SSA** can give two triangles; **AAA** fixes only the shape.","Regular n-gon: **central angle 360° ÷ n**, **interior angle (n − 2) × 180° ÷ n**, **mitre cut 180° ÷ n**.","Hexagon: step the radius six times. Octagon and dodecagon: bisect the square's and hexagon's central angles. Pentagon (72°): Euclid IV.11.","**Gauss (1796):** the 17-gon is constructible. A regular n-gon is constructible exactly when n = a power of 2 × distinct Fermat primes (3, 5, 17, 257, 65,537).","From 3 to 20, constructible: 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20. Not: 7, 9, 11, 13, 14, 18, 19.","**Trisection is impossible in general** (Wantzel, 1837): 20° cannot be constructed. Archimedes' marked ruler and origami can trisect, by breaking the classical rules.","**Whole-degree angles:** exactly the **multiples of 3°** are constructible (72° − 60° = 12° → 6° → 3°).","Chord-trisection is only approximate: for 60° it gives about 19.1°, 21.8°, 19.1°.","Heights by angle: at 45° elevation, height above eye = distance. A 3-4-5 rope makes a right angle.","Clock hands: minute hand 6° per minute, hour hand 30° per hour + 0.5° per minute.",{"id":1057,"type":1058,"sourceIds":1059},"sources-extend","sources",[1060,1061,1062,1063,1064,1065,1066,1067,1068,1069,1070,1071,1072,1073],"constructing-angles-ncert-math-6-practical-geometry","constructing-angles-ncert-ganita-prakash-6","constructing-angles-ncert-math-7-practical-geometry","constructing-angles-mathsisfun-protractor","constructing-angles-mathsisfun-constructions","constructing-angles-mathsisfun-degrees","constructing-angles-wikipedia-straightedge-compass","constructing-angles-wikipedia-angle-trisection","constructing-angles-wikipedia-shulba-sutras","constructing-angles-britannica-euclid-elements","constructing-angles-wikipedia-exact-trig-values","constructing-angles-wikipedia-paper-folding","constructing-angles-sciam-heptadecagon","constructing-angles-wikipedia-squaring-circle",[1060,1061,1062,1063,1064,1065,1066,1067,1068,1069,1070,1071,1072,1073],"needs_review",{"generatedBy":1077,"notes":1078},"claude-code","Draft generated with Python; every angle fact was computed and asserted. Pending owner review.","98a9e7a906bc3c2b2b07ba9bbabaef695371fe36a7a921186b2bf80719ca2613",{"component:match-pairs@1":1081,"logic:practice":1082,"component:compass-construction@1":1083,"component:sort-game@1":1084,"component:protractor@1":1085,"component:angle-lab@1":1086,"source:constructing-angles-britannica-euclid-elements":1087,"source:constructing-angles-mathsisfun-constructions":1088,"source:constructing-angles-mathsisfun-degrees":1089,"source:constructing-angles-mathsisfun-protractor":1090,"source:constructing-angles-ncert-ganita-prakash-6":1091,"source:constructing-angles-ncert-math-6-practical-geometry":1092,"source:constructing-angles-ncert-math-7-practical-geometry":1093,"source:constructing-angles-sciam-heptadecagon":1094,"source:constructing-angles-wikipedia-angle-trisection":1095,"source:constructing-angles-wikipedia-exact-trig-values":1096,"source:constructing-angles-wikipedia-paper-folding":1097,"source:constructing-angles-wikipedia-shulba-sutras":1098,"source:constructing-angles-wikipedia-squaring-circle":1099,"source:constructing-angles-wikipedia-straightedge-compass":1100},"2a8ee4ac87460b4e1175a4bb13c96b03d577db06dde95670eb7fcfe4ad787899","3d6b0fe1b15255975a32b0fcd94e8019bc959ad45cbf12e136e86149549c6878","0eba81381f31d8d78008911eb4c3d62745efd33b4ffadd94aef0851d0a2ad5f3","b164f45a2c8ca08f26c450768ff0231e113e9fe45381eddb34dc6d0548596c38","7407db21456592711dd16c6bdad23f042e85ebab9c314b072b17b7065eaf8ae3","1916502bd0021560b90784e9e612bc92ea263e6ed383d55cff0b75753b92213f","ad1a9a6a227fda5d3c1569f37efbe35e448ebaceba8cba872821fd48e2e00ed6","b93faffbd9c4d40f5fce2bc4b2ea0ab5ac64bb8c176f5e2bba3f37444df5e400","216eb0db510461864a47157f14054a39e15b1b0fc461b0fbc77664d9eb28b91d","4d3f50c07f44df57c80455dc39e01b2aa11bb0ee40811fca3b0f12d16e0b3f5e","acad5a4d56d24a5c1ad6e908f3809f2e7b3978f7c2810a9b33cdf82a65c4bb47","4aedaa1be389589b6e840923ef4e92fd15d03eda0b0ba0302d57b7e71bcb3dc7","21119e12648b9efd4cc82b11c59d626f2a53eace3a70552041f26c311b77ba2d","2ab885aef682a4c817268dd75b6110889b7897fc18b98ebc081e4aa62b5419f7","7f00387dc29d17172d25b6aa96420e2544a8bc59edf939af3dce91d515300a1f","fb92459a92df88915d8e44ab46afef7a382dc86bcdfa71f34f9c39dee944ab73","9e200fdcd862eae17818b04bb75d0d8ee8c56cbf629dc92b28a82ab307604f76","f7559697a2f963f9cb1e02a93fc5697840f583f22605d7483d688664862d70f9","d66ae780e95eb677a3ae7bcb24aa2430951fe7dded13b8366b75578ca4089dee","90d6cc756bc1bdd6cde0d5e4ed2000c88c3e2f3a8d91fdaf0ec4ab73af72b434",{"state":1102,"reviewer":1103,"selfReview":746,"reviewedAt":1104,"method":1105},"approved","The library owner","2026-09-20T10:18:37.581Z","owner_bulk","preview-7e1cbbcc4f",1789899597632]