Algebra & Trigonometry
The x + 1/x Family — the Paper’s Favourite Pattern
No single pattern appears more often in the exam algebra than this one. You are given x + 1/x = k and asked for x² + 1/x², x³ + 1/x³, x⁴ + 1/x⁴ or x − 1/x. Every one of them follows from squaring or cubing, and the beauty is that the cross terms always cancel: when you square x + 1/x you get x² + 2·x·(1/x) + 1/x² = x² + 1/x² + 2, because x × 1/x = 1. That single fact generates the whole chain.
| Asked for | Formula in k | k = 3 | k = 4 | k = 5 |
|---|---|---|---|---|
| x² + 1/x² | k² − 2 | 7 | 14 | 23 |
| x³ + 1/x³ | k³ − 3k | 18 | 52 | 110 |
| x⁴ + 1/x⁴ | (k² − 2)² − 2 | 47 | 194 | 527 |
| x − 1/x | ±√(k² − 4) | ±√5 | ±√12 = ±2√3 | ±√21 |
| x² − 1/x² | ±k√(k² − 4) | ±3√5 | ±8√3 | ±5√21 |
| x⁵ + 1/x⁵ | (k² − 2)(k³ − 3k) − k | 123 | 724 | 2525 |
Where does each row come from? Square the given: (x + 1/x)² = x² + 1/x² + 2 = k², so x² + 1/x² = k² − 2. Cube it: (x + 1/x)³ = x³ + 1/x³ + 3(x + 1/x), so x³ + 1/x³ = k³ − 3k. Square the second result: (x² + 1/x²)² = x⁴ + 1/x⁴ + 2, so x⁴ + 1/x⁴ = (k² − 2)² − 2. And since (x − 1/x)² = x² + 1/x² − 2 = (k² − 2) − 2 = k² − 4, we get x − 1/x = ±√(k² − 4). Nothing here is memorised blindly — each line is one squaring away from the line above.
- Worked example 1. x + 1/x = 3. Then x² + 1/x² = 9 − 2 = 7; x³ + 1/x³ = 27 − 9 = 18; x⁴ + 1/x⁴ = 7² − 2 = 47; x − 1/x = ±√(9 − 4) = ±√5. Numerical check: x + 1/x = 3 gives x = (3 + √5)/2 ≈ 2.618034, so 1/x ≈ 0.381966. Then x² + 1/x² ≈ 6.854 + 0.146 = 7.000 ✓ and x − 1/x ≈ 2.618 − 0.382 = 2.236 = √5 ✓
- Worked example 2 (the surd disguise). If x = 2 + √3, find x³ + 1/x³. First note 1/x = 1/(2 + √3) = (2 − √3)/[(2 + √3)(2 − √3)] = (2 − √3)/(4 − 3) = 2 − √3. So x + 1/x = (2 + √3) + (2 − √3) = 4 — the surds vanish. Now k = 4 gives x³ + 1/x³ = 4³ − 3(4) = 64 − 12 = 52. Numerical check: x ≈ 3.732051, x³ ≈ 51.9808, 1/x³ ≈ 0.019238, sum ≈ 52.000 ✓
- Worked example 3. If x = 3 + 2√2 then 1/x = 3 − 2√2 (since (3 + 2√2)(3 − 2√2) = 9 − 8 = 1), so x + 1/x = 6. Hence x² + 1/x² = 36 − 2 = 34 and x³ + 1/x³ = 216 − 18 = 198.
- Worked example 4 (given the square, work backwards). If x² + 1/x² = 7, find x³ + 1/x³. From k² − 2 = 7 we get k² = 9, so x + 1/x = 3 (take the positive value), and then x³ + 1/x³ = 27 − 9 = 18.
- Worked example 5 (the minus version). If x − 1/x = 3, then squaring gives x² + 1/x² = 3² + 2 = 11 (note the plus), and cubing gives x³ − 1/x³ = 3³ + 3(3) = 36. Also x + 1/x = ±√(9 + 4) = ±√13. The rule is simple: with the minus sign, every 2 and 3 that was subtracted is now added.
Linear Equations in One and Two Variables
A linear equation has every variable to the first power only. In one variable, ax + b = 0 always has exactly one root, x = −b/a (provided a ≠ 0). In two variables there are two standard methods — substitution (make one variable the subject and put it into the other equation) and elimination (multiply to make one coefficient match, then add or subtract). Elimination is faster in the exam because you avoid fractions.
- Worked example (substitution). Solve 2x + 3y = 13 and 3x − y = 3. From the second, y = 3x − 3. Substitute: 2x + 3(3x − 3) = 13 → 2x + 9x − 9 = 13 → 11x = 22 → x = 2, so y = 3(2) − 3 = 3. Check both: 2(2) + 3(3) = 4 + 9 = 13 ✓ and 3(2) − 3 = 3 ✓
- Worked example (elimination). Same pair. Multiply the second equation by 3: 9x − 3y = 9. Add it to the first: (2x + 3y) + (9x − 3y) = 13 + 9 → 11x = 22 → x = 2, then y = 3. Identical answer, no fractions on the way.
- Worked example (word problem). The sum of two numbers is 45 and their difference is 11. Adding the two statements: 2 × (larger) = 56, so the larger is 28 and the smaller is 45 − 28 = 17. Check: 28 + 17 = 45 ✓ and 28 − 17 = 11 ✓ Shortcut worth memorising: larger = (sum + difference)/2, smaller = (sum − difference)/2.
- Worked example (ages). A father is 3 times as old as his son. After 12 years he will be twice as old. Let the son be s, so the father is 3s. Then 3s + 12 = 2(s + 12) → 3s + 12 = 2s + 24 → s = 12 and the father is 36. Check: in 12 years they will be 24 and 48, and 48 = 2 × 24 ✓
| Condition | Number of solutions | The two lines are… | Example |
|---|---|---|---|
| a₁/a₂ ≠ b₁/b₂ | Exactly one (consistent, independent) | Intersecting at one point | 2x + 3y = 13, 3x − y = 3 → (2, 3) |
| a₁/a₂ = b₁/b₂ = c₁/c₂ | Infinitely many (dependent) | Coincident — the same line twice | 2x + 3y = 7 and 4x + 6y = 14 |
| a₁/a₂ = b₁/b₂ ≠ c₁/c₂ | No solution (inconsistent) | Parallel and distinct | 2x + 3y = 7 and 4x + 6y = 15 |
Worked example (find k). For what value of k does the pair kx + 3y = k − 3 and 12x + ky = k have no solution? Parallel lines need a₁/a₂ = b₁/b₂, so k/12 = 3/k → k² = 36 → k = 6 or −6. Now test each. With k = 6 the equations are 6x + 3y = 3 and 12x + 6y = 6 — the second is exactly twice the first, so there are infinitely many solutions, not none. With k = −6 they are −6x + 3y = −9 and 12x − 6y = −6; dividing the first by −3 gives 2x − y = 3 while dividing the second by 6 gives 2x − y = −1, a flat contradiction. Hence k = −6. The lesson: after using the ratio condition you must still eliminate the value that makes the constants match too.
Quadratic Equations — Roots, Discriminant, Sum and Product
A quadratic is ax² + bx + c = 0 with a ≠ 0. Three tools solve every exam question on it: factorisation by splitting the middle term, the quadratic formula x = [−b ± √(b² − 4ac)] / 2a, and the sum-and-product relations. Learn to see which one the question wants. If the question asks for the roots, factorise; if it asks about the nature of the roots, compute only the discriminant; if it asks for a symmetric expression in the roots such as α² + β², never find the roots at all — use sum and product.
- Sum of roots α + β = −b/a. Product of roots αβ = c/a. These two are the highest-yield facts in the whole topic.
- Forming the equation from roots: x² − (sum)x + (product) = 0. So the equation with roots 3 and −5 is x² − (−2)x + (−15) = x² + 2x − 15 = 0. Check by the formula: x = [−2 ± √(4 + 60)]/2 = (−2 ± 8)/2 = 3 and −5 ✓
- Reciprocal roots: if ax² + bx + c = 0 has roots α and β, then the equation with roots 1/α and 1/β is cx² + bx + a = 0 — simply swap a and c. Roots with signs flipped (−α, −β): ax² − bx + c = 0.
- Worked example (factorisation). x² − 7x + 12 = 0. Find two numbers with product 12 and sum 7: they are 3 and 4. So (x − 3)(x − 4) = 0 and the roots are 3 and 4. Cross-check: sum = 7 = −(−7)/1 ✓, product = 12 = 12/1 ✓
- Worked example (formula). 2x² − 5x − 3 = 0. Here b² − 4ac = 25 − 4(2)(−3) = 25 + 24 = 49, and √49 = 7. So x = (5 ± 7)/4, giving x = 3 and x = −1/2. Check: sum = 3 − 0.5 = 2.5 = −(−5)/2 ✓, product = 3 × (−0.5) = −1.5 = −3/2 ✓
- Worked example (symmetric expressions). If α and β are the roots of x² − 5x + 6 = 0, then α + β = 5 and αβ = 6. So α² + β² = 5² − 2(6) = 13; 1/α + 1/β = (α + β)/αβ = 5/6; α³ + β³ = 5³ − 3(6)(5) = 125 − 90 = 35; and (α − β)² = 25 − 24 = 1 so α − β = ±1. Verify with the actual roots 2 and 3: 4 + 9 = 13 ✓, 8 + 27 = 35 ✓, 3 − 2 = 1 ✓
| Value of D | Nature of the roots | Worked example |
|---|---|---|
| D > 0 and a perfect square (a, b, c rational) | Two distinct rational roots | 6x² − 5x + 1 = 0: D = 25 − 24 = 1, roots 1/2 and 1/3 |
| D > 0 but not a perfect square | Two distinct irrational roots, always a conjugate pair p ± √q | x² − 4x + 1 = 0: D = 16 − 4 = 12, roots 2 ± √3 |
| D = 0 | Roots real and equal, both = −b/2a | x² − 6x + 9 = 0: D = 36 − 36 = 0, both roots 3 |
| D < 0 | No real roots (a complex conjugate pair) | x² + x + 1 = 0: D = 1 − 4 = −3 |
Nature-of-roots question. For what positive k does x² − kx + 9 = 0 have equal roots? Equal roots need D = 0, so k² − 4(1)(9) = 0 → k² = 36 → k = ±6, and the positive value is 6. Then the equation is x² − 6x + 9 = (x − 3)², with both roots 3. Common-root question. Find k if x² − 5x + 6 = 0 and x² + kx − 6 = 0 share a root. The first has roots 2 and 3. Put x = 2 in the second: 4 + 2k − 6 = 0 → k = 1; put x = 3: 9 + 3k − 6 = 0 → k = −1. Verify k = 1: x² + x − 6 = (x + 3)(x − 2), roots −3 and 2, and 2 is shared ✓ Verify k = −1: x² − x − 6 = (x − 3)(x + 2), roots 3 and −2, and 3 is shared ✓
Polynomials, Surds, Indices and Algebraic Fractions
The remainder theorem says: the remainder when a polynomial p(x) is divided by (x − a) is simply p(a). The factor theorem is its special case: (x − a) is a factor of p(x) exactly when p(a) = 0. Together they replace long division with a single substitution — the biggest time saving in the whole algebra section. For a divisor of the form (bx − a), evaluate at x = a/b.
- Worked example 1. Remainder when x³ + 3x² − 5x + 4 is divided by (x − 2) = p(2) = 8 + 12 − 10 + 4 = 14.
- Worked example 2. Is (x − 3) a factor of x³ − 3x² + 4x − 12? p(3) = 27 − 27 + 12 − 12 = 0, so yes. Dividing out gives x³ − 3x² + 4x − 12 = (x − 3)(x² + 4), and since x² + 4 never vanishes for real x, the only real root is x = 3.
- Worked example 3 (find the constant). If (x − 2) is a factor of x³ − kx² + 2x − 8, then p(2) = 8 − 4k + 4 − 8 = 0 → 4k = 4 → k = 1. Check with k = 1: 8 − 4 + 4 − 8 = 0 ✓
- Worked example 4 (fractional divisor). Remainder when x³ − 2x² + 3x − 1 is divided by (2x − 3) = p(3/2) = 27/8 − 2(9/4) + 9/2 − 1 = 27/8 − 36/8 + 36/8 − 8/8 = 19/8.
Surds. To rationalise a denominator, multiply above and below by the conjugate. So 1/(√5 − √3) = (√5 + √3)/[(√5)² − (√3)²] = (√5 + √3)/(5 − 3) = (√5 + √3)/2. And 1/(3 + 2√2) = (3 − 2√2)/(9 − 8) = 3 − 2√2. A very common the paper step is to un-square a nested surd: √(7 + 4√3) = 2 + √3, because (2 + √3)² = 4 + 4√3 + 3 = 7 + 4√3 ✓ Likewise √(11 − 6√2) = 3 − √2, since (3 − √2)² = 9 − 6√2 + 2 = 11 − 6√2 ✓ The recipe: write the middle term 4√3 as 2 × 2 × √3, so the two parts are 2 and √3.
Indices. The laws are a^m × a^n = a^(m+n), a^m ÷ a^n = a^(m−n), (a^m)^n = a^(mn), a⁰ = 1, a^(−n) = 1/a^n and a^(1/n) = ⁿ√a. The exam technique is always the same: make the bases equal, then equate the powers. Worked example: solve 2^(2x−1) = 8^(x−3). Write 8 as 2³, so the right side is 2^(3x−9). Now 2x − 1 = 3x − 9 → x = 8. Check: 2^15 = 32768 and 8^5 = 32768 ✓ Another: 3^(x+1) = 81 = 3⁴ → x + 1 = 4 → x = 3. A classic identity question: if xᵃ = y, yᵇ = z and z^c = x, then x^(abc) = x, so abc = 1.
Algebraic fractions. Always factorise first, cancel second. Example: (x² − 9)/(x² + x − 12) = [(x − 3)(x + 3)]/[(x + 4)(x − 3)] = (x + 3)/(x + 4) for x ≠ 3. Test at x = 5: left side = (25 − 9)/(25 + 5 − 12) = 16/18 = 8/9; right side = 8/9 ✓ Another standard step: 1/(x − 1) − 1/(x + 1) = [(x + 1) − (x − 1)]/(x² − 1) = 2/(x² − 1). And a cyclic expression the paper likes to set: a/[(a − b)(a − c)] + b/[(b − a)(b − c)] + c/[(c − a)(c − b)] = 0 for all distinct a, b, c. Test with a = 1, b = 2, c = 3: 1/[(−1)(−2)] + 2/[(1)(−1)] + 3/[(2)(1)] = 0.5 − 2 + 1.5 = 0 ✓
Graphs of Linear Equations and Coordinate Geometry
Every linear equation in two variables is a straight line, and every straight line is a linear equation. The most useful form is y = mx + c, where m is the slope (how much y rises per unit rise in x) and c is the y-intercept. For the general form ax + by + c = 0: slope = −a/b, x-intercept = −c/a (set y = 0), y-intercept = −c/b (set x = 0). Worked example: 2x + 3y = 12 has slope −2/3; setting y = 0 gives x-intercept 6, setting x = 0 gives y-intercept 4. The line therefore cuts the axes at (6, 0) and (0, 4), and the triangle it forms with the axes has area ½ × 6 × 4 = 12 square units.
| Quantity | Formula | Worked value |
|---|---|---|
| Slope through two points | m = (y₂ − y₁)/(x₂ − x₁) | (1, 2) and (4, 6): m = 4/3 |
| Distance between two points | √[(x₂ − x₁)² + (y₂ − y₁)²] | (1, 2) to (4, 6): √(9 + 16) = 5 |
| Midpoint of a segment | ((x₁ + x₂)/2, (y₁ + y₂)/2) | (2, 3) and (8, 11) → (5, 7) |
| Section formula (ratio m : n) | ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) | Midpoint is the case m : n = 1 : 1 |
| Centroid of a triangle | ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3) | (0,0), (6,0), (3,9) → (3, 3) |
| Area of a triangle from vertices | ½ |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)| | (1,2), (4,6), (7,2) → 12 |
| Parallel lines | m₁ = m₂ (equal slopes) | y = 2x + 1 ∥ y = 2x − 7 |
| Perpendicular lines | m₁ × m₂ = −1 | 3x + 4y = 7 has m = −3/4, so ⊥ slope = 4/3 |
| Line through a point with slope m | y − y₁ = m(x − x₁) | (2, 3) with m = 2 → y = 2x − 1 |
- Worked example 1 (area). Vertices (1, 2), (4, 6), (7, 2). Area = ½ |1(6 − 2) + 4(2 − 2) + 7(2 − 6)| = ½ |4 + 0 − 28| = ½ × 24 = 12. Sanity check without the formula: the side from (1, 2) to (7, 2) is horizontal with length 6, and the third vertex is 4 units above that line, so area = ½ × 6 × 4 = 12 ✓
- Worked example 2 (collinearity). Are (1, 1), (2, 3) and (3, 5) collinear? Area = ½ |1(3 − 5) + 2(5 − 1) + 3(1 − 3)| = ½ |−2 + 8 − 6| = 0, so yes — three points are collinear exactly when the triangle they form has zero area. Equivalently, the slope from the first to the second is (3 − 1)/(2 − 1) = 2 and from the second to the third is (5 − 3)/(3 − 2) = 2 — equal, so collinear ✓
- Worked example 3 (right angle). The triangle (0, 0), (4, 0), (0, 3) has legs along the axes of length 4 and 3, so its area = ½ × 4 × 3 = 6 and its hypotenuse = √(16 + 9) = 5. The formula agrees: ½ |0(0 − 3) + 4(3 − 0) + 0(0 − 0)| = ½ × 12 = 6 ✓
Trigonometric Ratios, Standard Angles and Identities
Take a right-angled triangle and fix one acute angle θ. The side facing θ is the perpendicular (opposite), the side next to θ (not the hypotenuse) is the base (adjacent), and the longest side is the hypotenuse. Then sin θ = perpendicular/hypotenuse, cos θ = base/hypotenuse, tan θ = perpendicular/base, and the other three are their reciprocals: cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ. Note carefully that sec pairs with cos and cosec pairs with sin — the names are deliberately crossed and this is the single most common slip.
| Ratio | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin θ | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos θ | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan θ | 0 | 1/√3 | 1 | √3 | not defined |
| cosec θ | not defined | 2 | √2 | 2/√3 | 1 |
| sec θ | 1 | 2/√3 | √2 | 2 | not defined |
| cot θ | not defined | √3 | 1 | 1/√3 | 0 |
Three fundamental identities generate almost every simplification: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ. Their rearranged forms are what you actually use: sin²θ = 1 − cos²θ; sec²θ − tan²θ = 1, so (sec θ − tan θ)(sec θ + tan θ) = 1 — the two are reciprocals of each other; and cosec²θ − cot²θ = 1, so (cosec θ − cot θ)(cosec θ + cot θ) = 1. Worked example: if sec θ + tan θ = 2, then sec θ − tan θ = 1/2 immediately, so adding gives 2 sec θ = 2.5 → sec θ = 1.25 and subtracting gives 2 tan θ = 1.5 → tan θ = 0.75. Check: sec²θ − tan²θ = 1.5625 − 0.5625 = 1 ✓
- Worked example 1 (build the triangle). If sin θ = 3/5, then the perpendicular is 3 and the hypotenuse 5, so the base = √(25 − 9) = 4. Hence cos θ = 4/5, tan θ = 3/4, sec θ = 5/4 and cosec θ = 5/3. Then sec θ + tan θ = 5/4 + 3/4 = 2, and indeed sec θ − tan θ = 5/4 − 3/4 = 1/2 = 1/2 ✓
- Worked example 2. If tan θ = 3/4 then sin θ = 3/5 and cos θ = 4/5, so (sin θ + cos θ)/(sin θ − cos θ) = (7/5)/(−1/5) = −7. Notice the sign — sin θ is smaller than cos θ here, so the denominator is negative.
- Worked example 3 (sum given, product asked). If sin θ + cos θ = 7/5, square both sides: 1 + 2 sin θ cos θ = 49/25 → 2 sin θ cos θ = 24/25 → sin θ cos θ = 12/25. Then (sin θ − cos θ)² = 1 − 24/25 = 1/25, so sin θ − cos θ = ±1/5. The pair really is 4/5 and 3/5: their sum is 7/5 ✓ and product 12/25 ✓
- Worked example 4 (pure table work). Evaluate tan²60° + 4cos²45° + 3sec²30° + 5cos²90°. Substituting: (√3)² + 4(1/√2)² + 3(2/√3)² + 5(0)² = 3 + 4(1/2) + 3(4/3) + 0 = 3 + 2 + 4 = 9.
- Worked example 5. sin 30° cos 60° + cos 30° sin 60° = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1. (Those who know the compound-angle formula will recognise this as sin(30° + 60°) = sin 90° = 1 — but the table alone is enough.)
Complementary Angles, Quadrant Signs and Maximum Values
Two angles are complementary when they add to 90°. Replacing θ by (90° − θ) swaps each ratio with its co-ratio: sin becomes cos, tan becomes cot, sec becomes cosec (and back again). This is why the whole family is named with "co-" prefixes. The paper exploits it constantly: any expression mixing an angle and its complement collapses to 0, 1 or a table value.
| Conversion | Classic exam expression | Value |
|---|---|---|
| sin(90° − θ) = cos θ | sin²20° + sin²70° = sin²20° + cos²20° | 1 |
| cos(90° − θ) = sin θ | cos 1° cos 2° … cos 90° (contains cos 90°) | 0 |
| tan(90° − θ) = cot θ | tan 10° tan 20° tan 70° tan 80° | 1 (each pair tan θ · cot θ = 1) |
| cot(90° − θ) = tan θ | tan 1° tan 2° … tan 89° | 1 (44 pairs, with tan 45° = 1 left in the middle) |
| sec(90° − θ) = cosec θ | sin θ · sec(90° − θ) = sin θ · cosec θ | 1 |
| cosec(90° − θ) = sec θ | cos θ · cosec(90° − θ) = cos θ · sec θ | 1 |
| Mixed use | sin 35° cos 55° + cos 35° sin 55° | 1 (cos 55° = sin 35°, sin 55° = cos 35°, so it is sin²35° + cos²35°) |
Why does tan 1° tan 2° … tan 89° equal 1? Pair the first with the last: tan 1° × tan 89° = tan 1° × cot 1° = 1. Then tan 2° × tan 88° = 1, and so on. There are 44 such pairs (1 with 89, 2 with 88, …, 44 with 46), and the leftover middle term is tan 45° = 1. So the product is 1⁴⁴ × 1 = 1. The same pairing logic answers any "product of tangents" question in five seconds — just check whether the angles pair up to 90°.
| Expression | Maximum | Minimum | Worked instance |
|---|---|---|---|
| a sin θ + b cos θ | +√(a² + b²) | −√(a² + b²) | 3 sin θ + 4 cos θ: max 5, min −5; 5 sin θ + 12 cos θ: max 13 |
| sin θ · cos θ | 1/2 (at θ = 45°) | −1/2 | Because sin θ cos θ = ½ sin 2θ and sin 2θ lies in [−1, 1] |
| sin⁴θ + cos⁴θ | 1 | 1/2 (at θ = 45°) | Equals 1 − 2sin²θcos²θ = 1 − ½sin²2θ |
| sin⁶θ + cos⁶θ | 1 | 1/4 (at θ = 45°) | Equals 1 − 3sin²θcos²θ = 1 − ¾sin²2θ |
| a²sec²θ + b²cosec²θ | unbounded | (a + b)² | 4sec²θ + 9cosec²θ: min = (2 + 3)² = 25 |
| a²tan²θ + b²cot²θ | unbounded | 2ab | 9tan²θ + 4cot²θ: min = 2 × 3 × 2 = 12 |
Heights and Distances — the Tower Problems
This is the one application of trigonometry that the paper asks in almost every shift. The angle of elevation is measured upward from the horizontal at the observer’s eye; the angle of depression is measured downward from the horizontal. Because the observer’s horizontal line and the ground are parallel, the angle of depression from the top of a tower to a point on the ground equals the angle of elevation of the tower’s top from that point — alternate angles. Almost every problem then reduces to one relation: tan(angle) = height ÷ horizontal distance. Take √3 = 1.732 and √2 = 1.414 throughout.
- Problem 1 (the one-step case). From a point on the ground 30 m from the foot of a tower, the top is seen at an elevation of 60°. Find the height. Configuration: right-angled triangle with the vertical tower AB (A at the foot, B at the top), the horizontal ground AP = 30 m, and the angle at P equal to 60°. Then tan 60° = AB/AP → √3 = h/30 → h = 30√3 ≈ 51.96 m. If the angle had been 45°, h would equal 30 m exactly; if 30°, h = 30/√3 = 10√3 ≈ 17.32 m. Rule of thumb: 45° means height = distance; 60° means height is √3 times the distance; 30° means height is the distance divided by √3.
- Problem 2 (the walk-towards case, 30° → 45°). The elevation of a tower’s top from a point P is 30°. Walking 40 m directly towards the tower to a point Q, the elevation becomes 45°. Find the height. Configuration: the foot A, then Q, then P all lie on one straight horizontal line with A nearest, and PQ = 40 m. Let h be the height. From Q: tan 45° = h/AQ → AQ = h. From P: tan 30° = h/AP → AP = h√3. Since AP − AQ = PQ: h√3 − h = 40 → h(√3 − 1) = 40 → h = 40/(√3 − 1). Rationalise: h = 40(√3 + 1)/[(√3)² − 1] = 40(√3 + 1)/2 = 20(√3 + 1) ≈ 54.64 m. Check: AQ = 54.64 and AP = 54.64 × 1.732 = 94.64, and 94.64 − 54.64 = 40 ✓
- Problem 3 (the shadow case, 30° and 60°). The shadow of a tower on level ground is 40 m longer when the sun’s altitude is 30° than when it is 60°. Find the height. The sun’s altitude is the angle of elevation of the sun, so tan(altitude) = height ÷ shadow, i.e. shadow = h/tan(altitude). Shadow at 30° = h√3; shadow at 60° = h/√3. Difference: h√3 − h/√3 = 40 → h(3 − 1)/√3 = 40 → 2h/√3 = 40 → h = 20√3 ≈ 34.64 m. Check: shadow at 30° = 20√3 × √3 = 60 m, shadow at 60° = 20√3/√3 = 20 m, difference = 40 m ✓ Useful special results: if the shadow equals the height, the altitude is 45°; if the shadow is √3 times the height, the altitude is 30°; if the shadow is 1/√3 of the height, the altitude is 60°.
- Problem 4 (two observers on opposite sides). Two men stand on opposite sides of a 100 m tower and see its top at elevations of 30° and 45°. How far apart are they? Configuration: the tower foot A is between the two men P and Q on one straight line. From the 30° man: AP = 100√3 ≈ 173.2 m. From the 45° man: AQ = 100 m. They are on opposite sides, so the distance PQ = AP + AQ = 100√3 + 100 = 100(√3 + 1) ≈ 273.2 m. Contrast with Problem 2, where both observers were on the same side and the distances were subtracted. Reading which side they stand on is the whole question.
- Problem 5 (two towers, elevation and depression together). From the top of a 7 m tall building the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower. Configuration: the building CD stands vertically (C on the ground, D on top, CD = 7 m); the tower AB stands vertically on the same level ground at horizontal distance d from the building; the horizontal line from D meets the tower at a point E, so DE = d and CE = 7 = AE. Depression of the foot A: tan 45° = AE/DE = 7/d → d = 7 m. Elevation of the top B: tan 60° = BE/DE = BE/7 → BE = 7√3. Total tower height AB = AE + EB = 7 + 7√3 = 7(1 + √3) ≈ 19.12 m.
- Problem 6 (across a river). Standing on one bank, a person sees the top of a tree on the opposite bank at an elevation of 60°. Moving 40 m straight back from the bank, the elevation becomes 30°. Find the height of the tree and the width of the river. Same configuration as Problem 2 but with the angles in the other order. Let h be the height and d the width (the horizontal distance from the near bank to the tree). Then tan 60° = h/d → d = h/√3, and tan 30° = h/(d + 40) → d + 40 = h√3. Subtracting: h√3 − h/√3 = 40 → 2h/√3 = 40 → h = 20√3 ≈ 34.64 m, and d = 20√3/√3 = 20 m. General shortcut for a 60° → 30° pair with walk-back distance d: h = d√3/2; here 40 × 1.732/2 = 34.64 ✓
(i) 30° → 45° gives h = d/(√3 − 1) = d(√3 + 1)/2;
(ii) 45° → 60° gives h = d/(1 − 1/√3) = d√3/(√3 − 1) = d(3 + √3)/2;
(iii) 30° → 60° gives h = d√3/2. Each follows from h·cot(smaller angle) − h·cot(larger angle) = d. Memorising the third one alone covers a large share of the questions, because 30° and 60° is the examiner’s favourite pair.
Key takeaways
- Learn a³ + b³ = (a + b)³ − 3ab(a + b) and a³ − b³ = (a − b)³ + 3ab(a − b): they turn "sum and product given, find the cubes" into one substitution.
- If a + b + c = 0 then a³ + b³ + c³ = 3abc. Scan every three-bracket question for a hidden zero sum before doing anything else.
- The x + 1/x chain: x² + 1/x² = k² − 2, x³ + 1/x³ = k³ − 3k, x⁴ + 1/x⁴ = (k² − 2)² − 2, x − 1/x = ±√(k² − 4). With a minus sign given instead, every subtraction becomes an addition.
- For a quadratic, sum of roots = −b/a and product = c/a; the equation from given roots is x² − (sum)x + (product) = 0. Never compute ugly roots for a symmetric expression.
- Discriminant D = b² − 4ac decides everything: D > 0 distinct real, D = 0 equal, D < 0 no real roots; and D > 0 with a perfect square means rational roots.
- Remainder theorem: the remainder on dividing p(x) by (x − a) is p(a); it is a factor exactly when p(a) = 0. One substitution replaces a full long division.
- Rebuild the standard-angle table from √(0/4), √(1/4), √(2/4), √(3/4), √(4/4) for sin; read it backwards for cos; divide for tan; invert for the other three.
- Any expression mixing θ with 90° − θ collapses: sin θ · cosec θ = 1, sin²20° + sin²70° = 1, tan 1° tan 2° … tan 89° = 1. Look for the pairing before you calculate.
- In heights and distances only one relation matters, tan(angle) = height ÷ distance. Same side of the tower means subtract the distances, opposite sides means add them; and 30° → 60° over a walk of d gives h = d√3/2.
Practice questions (8)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
If x + 1/x = 4, then the value of x² + 1/x² is:
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Answer: A — 14
Square the given expression: (x + 1/x)² = x² + 2(x)(1/x) + 1/x² = x² + 1/x² + 2, because x × 1/x = 1. So x² + 1/x² = 4² − 2 = 16 − 2 = 14. The general result is x² + 1/x² = k² − 2. The trap option 16 is what you get if you forget to subtract the 2.If a + b + c = 0 and abc = 5, then a³ + b³ + c³ equals:
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Answer: C — 15
The identity a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) has (a + b + c) as a factor on the right. When a + b + c = 0 the whole right side vanishes, leaving a³ + b³ + c³ = 3abc = 3 × 5 = 15. Test the rule on real numbers: 2 + (−5) + 3 = 0, and 8 − 125 + 27 = −90, while 3 × 2 × (−5) × 3 = −90 as well.If α and β are the roots of x² − 5x + 6 = 0, then α³ + β³ is:
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Answer: B — 35
Do not solve for the roots. Sum α + β = −(−5)/1 = 5 and product αβ = 6/1 = 6. Now α³ + β³ = (α + β)³ − 3αβ(α + β) = 125 − 3(6)(5) = 125 − 90 = 35. Confirmation: the roots really are 2 and 3, and 8 + 27 = 35. The sum-and-product route works even when the roots are irrational, which is why it is the method to train.For which positive value of k does the equation x² − kx + 9 = 0 have two equal roots?
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Answer: D — 6
Equal roots require the discriminant to be zero: b² − 4ac = 0. Here a = 1, b = −k, c = 9, so k² − 36 = 0 → k² = 36 → k = ±6, and the positive value is 6. Check: x² − 6x + 9 = (x − 3)², so both roots are 3. Option 3 is the trap for candidates who take the square root of 9 instead of setting the discriminant to zero.The value of sin²20° + sin²70° is:
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Answer: C — 1
The two angles are complementary, since 20° + 70° = 90°. So sin 70° = sin(90° − 20°) = cos 20°, and the expression becomes sin²20° + cos²20°, which is 1 by the fundamental identity. Spotting that a pair of angles adds to 90° is the fastest route in almost every the paper trigonometry-value question.Consider the following two statements about maximum values:
1. The maximum value of 5 sin θ + 12 cos θ is 13.
2. The maximum value of sin θ · cos θ is 1.
Which is/are correct?Show answer
Answer: A — Only 1
Statement 1 is correct: a sin θ + b cos θ ranges over ±√(a² + b²), and here √(25 + 144) = √169 = 13. Statement 2 is wrong: sin θ cos θ = ½ sin 2θ, and since sin 2θ can be at most 1, the maximum of sin θ cos θ is only 1/2, reached at θ = 45° where sin θ = cos θ = 1/√2 and the product is 1/2. So only statement 1 holds.The shadow of a tower on level ground is 20 m longer when the sun's altitude is 30° than when it is 60°. The height of the tower is: (take √3 = 1.732)
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Answer: D — 10√3 m
Shadow = height ÷ tan(altitude). At 30° the shadow is h/tan 30° = h√3; at 60° it is h/tan 60° = h/√3. The difference is h√3 − h/√3 = h(3 − 1)/√3 = 2h/√3 = 20, so h = 10√3 ≈ 17.32 m. Check: shadow at 30° = 10√3 × √3 = 30 m and at 60° = 10√3/√3 = 10 m, a difference of exactly 20 m. The general shortcut for a 30°/60° shadow pair with difference d is h = d√3/2.The area of the triangle whose vertices are (1, 2), (4, 6) and (7, 2) is:
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Answer: B — 12 square units
Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)| = ½ |1(6 − 2) + 4(2 − 2) + 7(2 − 6)| = ½ |4 + 0 − 28| = ½ × 24 = 12. Sanity check without the formula: (1, 2) and (7, 2) both lie on the line y = 2, so that side is horizontal with length 7 − 1 = 6, and the third vertex is 6 − 2 = 4 units above it, giving area = ½ × 6 × 4 = 12. Option 24 is what you get if you forget the ½.