Secondary Mathematics · Worked Repair Guide 06
Geometry becomes reliable when a diagram is treated as a record of stated relationships rather than as a picture to be trusted by eye. A line may look horizontal without being declared horizontal. Two angles may look equal without any information proving equality. A triangle may look isosceles without being one.
This guide develops the habit statement → reason → consequence. Every geometric conclusion should be tied to a fact, definition, theorem or calculation. That discipline connects angle work, similarity, congruence, Pythagoras, trigonometry and coordinate geometry.
The diagrams described here are original teaching constructions. Sketch them yourself where useful, marking only the information actually given.
1. Begin by separating given facts from visual appearance
Suppose a drawn angle appears to be 90°. Unless the problem states a right angle, shows an accepted right-angle mark, or supplies enough information to prove 90°, the appearance alone is not evidence.
A good annotation system uses one colour or symbol for given facts and another for conclusions. For example: “AB = AC” may be given; “∠ABC = ∠BCA” is then a consequence because base angles of an isosceles triangle are equal.
Entry check: a triangle is drawn with sides that look like 5 cm, 5 cm and 8 cm but no lengths are marked. Can you claim it is isosceles? No. The picture suggests a possibility but does not establish a mathematical condition.
2. Straight lines, points and vertically opposite angles
Angles on a straight line sum to 180°. Angles around a point sum to 360°. Vertically opposite angles formed by two intersecting straight lines are equal.
Worked example: two straight lines intersect. One angle is 68°. The vertically opposite angle is also 68°. Each adjacent angle is 180° − 68° = 112°.
The reason matters because several angle facts may produce the same arithmetic. Writing “112° because angles on a straight line sum to 180°” records which relationship was used.
When three or more rays meet at a point, identify exactly which angles partition the full turn. Do not add angles that overlap.
3. Parallel lines create angle relationships through a transversal
When parallel lines are cut by a transversal, corresponding and alternate angles are equal, while co-interior angles on the same side sum to 180°.
The parallel condition is essential. Without it, those angle relationships do not follow.
Worked example: two parallel lines are cut by a transversal. An acute angle at the first intersection is 57°. The corresponding acute angle at the second intersection is 57°. Its adjacent obtuse angle is 123°.
If a diagram shows arrow markings indicating parallel lines, copy that information into your reasoning before using a parallel-line fact. A student who remembers “Z angles are equal” without checking the arrows is using a visual slogan instead of a condition.
4. Triangle angle sums and exterior angles
The interior angles of a Euclidean triangle sum to 180°. Therefore if two angles are 43° and 71°, the third is 66°.
An exterior angle formed by extending one side equals the sum of the two opposite interior angles. This follows from combining the 180° straight-line sum with the triangle angle sum.
Worked derivation: if the exterior angle is E and its adjacent interior angle is C, then E + C = 180°. If A + B + C = 180°, subtracting C gives E = A + B.
Understanding the derivation is stronger than treating the exterior-angle result as an unrelated rule.
5. Isosceles triangles work in both directions—with conditions
If two sides of a triangle are equal, the angles opposite them are equal. Conversely, if two angles of a triangle are equal, the sides opposite them are equal.
Suppose AB = AC. Then ∠B = ∠C. If ∠A = 40°, the remaining 140° is shared equally, so ∠B = ∠C = 70°.
Do not conclude that a triangle is equilateral merely because it looks symmetrical. Three equal angles imply three equal sides; two equal angles imply only an isosceles relationship unless the third is also equal.
6. Polygon angle sums come from triangles
A convex n-sided polygon can be divided from one vertex into n − 2 triangles. Therefore its interior angle sum is (n − 2)×180°.
For a hexagon, the sum is 4×180° = 720°. If the hexagon is regular, each interior angle is 720°/6 = 120°.
The exterior angles of any convex polygon, taken once around in the same direction, sum to 360°. For a regular n-gon, each exterior angle is 360°/n.
Worked example: a regular polygon has exterior angle 24°. Then n = 360/24 = 15 sides.
7. Congruence and similarity are not the same claim
Congruent shapes have the same size and shape. Similar shapes have the same shape but may differ in size by a common scale factor.
If two triangles are congruent, corresponding lengths are equal and corresponding angles are equal. If they are similar, corresponding angles are equal and corresponding side lengths are proportional.
Similarity therefore requires consistent correspondence. A side of length 6 in one triangle cannot be paired with whichever side produces a convenient ratio.
Worked example: triangle ABC is similar to triangle DEF with A↔D, B↔E and C↔F. If AB = 6, DE = 9 and BC = 8, the scale factor from ABC to DEF is 9/6 = 1.5, so EF = 12.
8. Similarity scales length, area and volume differently
If corresponding lengths scale by k, areas scale by k². For similar solids, volumes scale by k³.
If two similar triangles have side-length factor 5/3, their area factor is 25/9. An area of 36 cm² in the smaller triangle corresponds to 36×25/9 = 100 cm² in the larger.
This follows because area contains two independent length dimensions. The measurement guide develops the same principle through unit conversion and scale drawings.
Do not add the percentage increase in length twice. Multiplicative scaling is handled by the square of the factor.
9. Pythagoras connects side lengths in a right triangle
For a right-angled triangle with legs a and b and hypotenuse c, a² + b² = c².
The hypotenuse is the side opposite the right angle. It is not simply the side that happens to be drawn diagonally.
Worked example: legs 9 cm and 12 cm give c² = 81 + 144 = 225, so c = 15 cm. A negative root is not a physical side length.
To find a leg, subtract: if c = 13 and a = 5, then b² = 169 − 25 = 144, so b = 12.
A quick check is that the hypotenuse must be longer than either leg and shorter than their sum.
10. Trigonometric ratios are side relationships tied to an angle
In a right triangle, relative to an acute angle θ, sine, cosine and tangent compare specific pairs of sides:
sin θ = opposite/hypotenuse
cos θ = adjacent/hypotenuse
tan θ = opposite/adjacent
The words opposite and adjacent depend on which angle is being considered. The hypotenuse does not.
Worked example: a right triangle has hypotenuse 10 cm and an acute angle 30°. The side opposite 30° is 10 sin 30° = 5 cm.
Before selecting a ratio, mark the known side, unknown side and reference angle. Then choose the ratio that contains exactly those sides.
11. Inverse trigonometric functions recover an angle
If tan θ = 3/4 for an acute angle in a right triangle, θ = tan⁻¹(3/4), approximately 36.9°.
The notation tan⁻¹ here means the inverse function used to recover an angle, not the reciprocal 1/tan θ. Context and calculator notation matter.
Check that the resulting angle is plausible. If the opposite side is smaller than the adjacent side, an acute angle below 45° is reasonable.
Use the calculator’s correct angle mode for the problem. A degree-mode question evaluated in radians can produce a numerically valid calculator output with the wrong interpretation.
12. Bearings and direction require a reference convention
Three-figure bearings are measured clockwise from north and written with three digits, such as 045°, 120° or 270°.
A bearing of 060° means rotate 60° clockwise from the north direction. It does not mean 60° from east.
The reverse bearing differs by 180° modulo 360°. Thus the reverse of 060° is 240°.
Worked example: B lies on a bearing 135° from A. Then A lies on a bearing 315° from B.
Draw north lines at the relevant points and mark the clockwise direction explicitly. Many bearing errors are reading errors before they are trigonometric errors.
13. Coordinate geometry can prove geometric claims
Suppose A = (0,0), B = (4,0) and C = (4,3). Segment AB is horizontal and BC vertical, so they are perpendicular. Their lengths are 4 and 3. The distance AC is √(4² + 3²) = 5.
The coordinate model therefore produces the familiar 3-4-5 right triangle. Geometry and algebra are not separate worlds; coordinates translate geometric relationships into numerical and symbolic ones.
For non-vertical lines, equal gradients indicate parallel lines. Gradients whose product is −1 indicate perpendicular non-vertical lines when both gradients exist.
Do not apply the gradient-product rule to a vertical line, whose gradient is undefined. Handle vertical and horizontal lines directly.
14. A proof is a chain, not a collection of true facts
In a geometric argument, each statement must contribute to the conclusion. Listing several correct angle facts does not automatically prove the desired result.
Suppose AB = AC and D lies on BC with AD perpendicular to BC. Triangles ABD and ACD have equal hypotenuses AB and AC, a common side AD and right angles at D. Under the appropriate right-triangle congruence condition, the triangles are congruent, so BD = DC.
The conclusion follows from the congruence, which follows from identified matching information. That chain is what makes the argument transferable.
If your course uses a specific accepted set of congruence criteria, state the relevant one accurately rather than writing only “triangles are the same”.
15. A capstone configuration
Triangle ABC is right-angled at B. Let AB = 6 cm, BC = 8 cm. Point D lies on AC such that AD:DC = 1:1. Find AC, the midpoint distance BD and angle A.
Pythagoras gives AC = √(6² + 8²) = 10 cm.
Since D is the midpoint of the hypotenuse of a right triangle, if that theorem is available in the learner’s course, BD = AC/2 = 5 cm. Alternatively, coordinates provide an independent route: set B=(0,0), A=(6,0), C=(0,8). Then D=(3,4), and BD=√(3²+4²)=5.
For angle A, opposite side BC = 8 and adjacent side AB = 6, so tan A = 8/6 = 4/3. Thus A ≈ 53.1°.
This one configuration links Pythagoras, midpoint reasoning, coordinates and trigonometry. Different representations reveal different features while preserving the same geometry.
16. Independent practice
- Two lines intersect and one angle is 74°. Find the other three angles.
- Angles on a straight line are x and 3x. Find x.
- A triangle has angles 42° and 68°. Find the third angle.
- An isosceles triangle has equal sides AB and AC and vertex angle A = 36°. Find B and C.
- Find the interior angle sum of an octagon.
- Find each exterior angle of a regular 12-gon.
- Similar triangles have corresponding sides 8 cm and 12 cm. Find the scale factor from smaller to larger.
- The smaller triangle in Question 7 has area 20 cm². Find the larger area.
- A right triangle has legs 7 cm and 24 cm. Find its hypotenuse.
- A right triangle has hypotenuse 17 cm and one leg 8 cm. Find the other leg.
- In a right triangle, the side opposite θ is 5 and the adjacent side is 12. Find θ.
- A 10 m ladder makes a 65° angle with the ground. Find the vertical height reached, ignoring ladder thickness.
- Find the reverse bearing of 028°.
- Find the reverse bearing of 215°.
- Points A=(1,2) and B=(5,10). Find the gradient AB.
- Line l has gradient 2. Find the gradient of a perpendicular non-vertical line.
- Two similar solids have length factor 3. Find their volume factor.
- Explain why a diagram that looks like a rectangle does not establish four right angles.
17. Worked answers
1. 74°, 106°, 74°, 106°. Vertically opposite angles are equal; adjacent angles on a straight line sum to 180°.
2. x = 45°. The equation is x + 3x = 180°.
3. 70°. Subtract the known angles from 180°.
4. 72° each. The equal base angles share 144°.
5. 1080°. Use (8−2)×180°.
6. 30°. Divide 360° by 12.
7. 1.5. Divide corresponding larger length by smaller length: 12/8.
8. 45 cm². Area factor = 1.5² = 2.25.
9. 25 cm. √(49+576)=25.
10. 15 cm. √(17²−8²)=√225.
11. Approximately 22.6°. θ = tan⁻¹(5/12).
12. Approximately 9.06 m. Height = 10 sin 65°.
13. 208°. Add 180°.
14. 035°. Subtract 180° from 215° and retain the three-figure format.
15. 2. Gradient = (10−2)/(5−1)=8/4.
16. −1/2. The gradients multiply to −1.
17. 27. Cube the length factor.
18. Appearance is not a stated condition. Right angles require markings, givens or a proof from other facts.
18. Repair geometric reasoning at the first unsupported claim
When a proof or angle solution fails, find the first statement whose reason is missing or false. Common causes include assuming parallel lines, pairing the wrong corresponding sides, choosing the wrong trigonometric sides, or trusting the scale of the sketch.
A good correction note names the exact relationship: “Used alternate angles without a parallel condition” or “Matched AB with DF instead of DE”. The next practice question should alter the diagram so that the same reasoning must be reconstructed rather than copied.
19. Continue through the BTT Mathematics library
Return to the BTT Mathematics Hub. Use Units, Scale and Measurement for dimensional and scale-factor work, Graphs, Tables and Relationships for coordinate geometry, and Signed Numbers, Brackets and Algebraic Structure for the algebra beneath geometric formulas.
The BTT Mathematical Lab is the diagnostic route when the same diagram-reading or theorem-selection error repeats.
20. Sources and scope
The angle, similarity, coordinate and trigonometric examples in this guide are original teaching constructions. The displayed calculations and justifications provide the mathematical support for the conclusions.
For the current Singapore Secondary curriculum doorway, see MOE: Curriculum for secondary schools. Match theorem names, notation and extension work to the learner’s actual school and subject level.