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Geometry and Trigonometry

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Geometry and Trigonometry

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Trigonometry in WASSCE focuses on right-angled triangles and the three primary ratios: sine (sin), cosine (cos), and tangent (tan). For a right-angled triangle with angle θ, opposite side O, adjacent side A, and hypotenuse H: sin θ = O/H, cos θ = A/H, tan θ = O/A. The mnemonic SOH-CAH-TOA helps you remember which sides go with which ratio.

Essential Formulas:

  • sin² θ + cos² θ = 1 (the Pythagorean identity)
  • tan θ = sin θ / cos θ
  • Area of triangle = ½ ab sin C (two sides a, b with included angle C)
  • Sine rule: a/sin A = b/sin B = c/sin C
  • Cosine rule: a² = b² + c² − 2bc cos A
  • Bearings measured clockwise from North (000°)
  • 180° − θ and 90° − θ trigonometric relationships

Key Facts:

  • Pythagorean triples: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25) — these appear frequently
  • sin 30° = cos 60° = ½; cos 30° = sin 60° = √3/2; tan 30° = 1/√3; tan 60° = √3
  • Angles of elevation and depression are equal (alternate angles with the horizontal)
  • Always draw a diagram for 3D geometry problems — label all known quantities

⚡ Exam Tip: WASSCE often asks for angles of elevation or depression. Draw a horizontal line from the observer’s eye to the object. The angle between this horizontal and the line of sight is the angle of elevation (if looking up) or depression (if looking down). For a problem where a building of height 20 m is observed at an angle of elevation 35°, use tan 35° = 20/d where d is the distance. So d = 20/tan 35° ≈ 20/0.700 ≈ 28.6 m.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Right-Angled Triangle Trigonometry

In the right-angled triangle ABC with right angle at C:

  • sin A = opposite/hypotenuse = BC/AB
  • cos A = adjacent/hypotenuse = AC/AB
  • tan A = opposite/adjacent = BC/AC

Example: A ladder 5 m long leans against a wall, making an angle of 65° with the ground. How far is the foot of the ladder from the wall? cos 65° = adjacent/5 = x/5 → x = 5 cos 65° = 5 × 0.4226 ≈ 2.11 m from the wall.

Angles of Elevation and Depression

The angle of elevation is the angle between the horizontal and the line of sight when looking UP at an object. The angle of depression is the angle between the horizontal and the line of sight when looking DOWN at an object. When the ground is level, these angles are equal because they are alternate angles.

Example: A man stands 50 m from a tree. From his eye level (1.6 m above ground), the angle of elevation to the top of the tree is 30°. Find the height of the tree. tan 30° = height difference / 50 → 1/√3 = h/50 → h = 50/√3 ≈ 28.87 m. Add the observer’s height: total tree height = 28.87 + 1.6 = 30.47 m.

The Sine Rule

Used when you know either: (a) two angles and one side, or (b) two sides and an angle opposite one of them. a/sin A = b/sin B = c/sin C

Example: In triangle ABC, angle A = 40°, angle B = 65°, and side a = 8 cm. Find side b. b/sin 65° = 8/sin 40° → b = 8 sin 65° / sin 40° = 8 × 0.9063 / 0.6428 ≈ 11.28 cm.

The Cosine Rule

Used when you know two sides and the included angle, or all three sides. a² = b² + c² − 2bc cos A. Rearranged: cos A = (b² + c² − a²) / 2bc.

Example: Find angle A in a triangle with sides a = 7 cm, b = 5 cm, c = 6 cm. cos A = (5² + 6² − 7²) / (2 × 5 × 6) = (25 + 36 − 49) / 60 = 12/60 = 0.2 → A = cos⁻¹(0.2) ≈ 78.5°.

Bearings

Bearings are always measured clockwise from North and written as three digits. A bearing of 045° is in the NE quadrant, 135° is SE, 225° is SW, 315° is NW.

Example: A ship travels 20 km on a bearing of 120° from point A. How far east and south has it travelled? East component: 20 sin 120° = 20 × 0.8660 = 17.32 km east. South component: 20 cos 120° = 20 × (−0.5) = 10 km south (negative means south).

Arc Length and Sector Area

For a circle of radius r and angle θ (in degrees):

  • Arc length = (θ/360) × 2πr
  • Sector area = (θ/360) × πr²

3D Trigonometry

Problems involving 3D shapes require identifying the correct right-angled triangles. Common approach: find or create right-angled triangles within the 3D figure. For a pyramid with a square base, dropping a perpendicular from the apex to the centre of the base creates a right-angled triangle with the slant height and vertical height.

Problem-Solving Strategies:

  • Always draw a diagram, even if one is provided — sketching helps you think
  • Label all given information on the diagram
  • Identify which formula applies: sine rule (AA or ASA/AAS), cosine rule (SAS or SSS), or SOH-CAH-TOA (right-angled)
  • When using the sine rule to find an angle, use the ambiguous case rule: sin θ = sin(180° − θ), so if sin A = 0.5, A could be 30° or 150° — check whether 150° is possible given the other angles
  • In bearing problems, draw North lines at key points to help visualise directions

Common Mistakes:

  • Using the wrong trigonometric ratio — always identify which sides are opposite and adjacent to the angle you’re working with
  • Forgetting to add or subtract the observer’s height when calculating total height from angle of elevation
  • Bearing errors: writing 45° instead of 045° (needs three digits); confusing clockwise and anticlockwise measurement
  • Mixing up when to use sine rule vs cosine rule — sine rule requires an angle and its opposite side; cosine rule is for SAS or SSS
  • In the ambiguous case (SSA), accepting only one solution when two are mathematically possible

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Proof of the Sine Rule

In any triangle ABC with altitude h from C to side AB: h = b sin A = a sin B. Therefore b/sin B = a/sin A. This extends to c/sin C, giving the complete sine rule: a/sin A = b/sin B = c/sin C = 2R, where R is the circumradius.

Proof of the Cosine Rule

Place triangle ABC with side BC along the x-axis. Coordinates: B = (0, 0), C = (a, 0), A = (b cos C, b sin C). By the distance formula: AB² = (b cos C)² + (b sin C)² = b² cos² C + b² sin² C = b². But also AB² = c². Therefore: c² = b² + a² − 2ab cos C. Replacing C with A gives the cosine rule.

Area Formula Derivation

The area of any triangle is ½ base × height. If we drop a perpendicular from C to side AB, the height h = b sin A. So area = ½ × c × (b sin A) = ½ ab sin C. This works for any two sides and the included angle.

The Ambiguous Case (SSA)

When given two sides a, b and a non-included angle A, the sine rule can give two possible solutions:

  • If a < b: no solution (a too short), one solution, or two solutions
  • If a = b sin A: exactly one solution (right triangle)
  • If b sin A < a < b: two solutions (the “crossed” configuration)
  • If a ≥ b: exactly one solution

Example: Given a = 8, b = 10, A = 30°. b sin A = 10 × ½ = 5. Since 5 < 8 < 10, there are two possible triangles. sin B = b sin A / a = 10 × ½ / 8 = 5/8 = 0.625. So B = sin⁻¹(0.625) = 38.68° or B = 180° − 38.68° = 141.32°. Both are possible since 30° + 141.32° < 180°.

Trigonometric Identities — Proving and Using

Fundamental identity: sin² θ + cos² θ = 1 Dividing by cos² θ: tan² θ + 1 = sec² θ Dividing by sin² θ: 1 + cot² θ = csc² θ

Example proof: Show that (sin θ + cos θ)² = 1 + 2 sin θ cos θ LHS = sin² θ + 2 sin θ cos θ + cos² θ = (sin² θ + cos² θ) + 2 sin θ cos θ = 1 + 2 sin θ cos θ. ✓

Solving Trigonometric Equations

Example: Solve 2 sin² x + cos x − 1 = 0 for 0° ≤ x ≤ 360°. Use sin² x = 1 − cos² x: 2(1 − cos² x) + cos x − 1 = 0 → 2 − 2 cos² x + cos x − 1 = 0 → −2 cos² x + cos x + 1 = 0 → 2 cos² x − cos x − 1 = 0 → (2 cos x + 1)(cos x − 1) = 0. cos x = 1 → x = 0°, 360°. cos x = −½ → x = 120°, 240°. All within range: x = 0°, 120°, 240°, 360°.

3D Applications — The Breadth and Depth of Problems

Example: A pyramid has a square base of side 8 cm and a height of 12 cm. Find the angle between a sloping edge and the base plane. The slant edge from apex to a base corner: in the base diagonal plane, half the diagonal = 4√2 cm. Using Pythagoras in 3D: slant edge² = 4² + 4² + 12² = 16 + 16 + 144 = 176 → slant edge = √176 = 2√44 cm. The line from the centre of the base to a corner = 4√2 cm (half diagonal). The angle φ between slant edge and base satisfies cos φ = (4√2) / (2√44) = (4√2) / (4√11) = √(2/11). So φ = cos⁻¹(√(2/11)) ≈ 57.7°.

WASSCE Exam Patterns

Paper 1: Multiple-choice questions testing direct substitution into sine/cosine/tan formulas, reading angles from diagrams, and simple identities. Paper 2: Typically one trigonometry question in Section A involving: (a) problem solving with angles of elevation/depression, (b) applying sine rule or cosine rule, (c) a trigonometric proof using identities.

⚡ Advanced Exam Tip: When a WASSCE question asks for “the angle between the line AB and the plane P”, find the angle between the line and its projection on the plane. If A is above the plane and B is on the plane, drop a perpendicular from A to the plane at point C. The angle is ∠ABC where AB is the sloping line and BC is on the plane. Use the sine or cosine rule in triangle ABC to find this angle.

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