Grade 11 — Term 3

Trigonometry

What you must know

• Angle of depression baseline trap: The angle of depression is measured from the HORIZONTAL line of sight, NEVER from the vertical cliff or mast.
• Observer height addition: If an observer of height h stands on ground looking at a building of height H, the opposite side is (H - h); remember to add h back when calculating total height.
• Trigonometric table lookups: In Paper 2 trigonometry problems, show the ratio substitution clearly before writing the four-figure table decimal reading.

Try one question

From the top of a 30-metre vertical cliff, the angle of depression of a boat out at sea is 30°. How far is the boat from the foot of the cliff? (Use tan 30° = 1/√3)

1. A. 30 m
2. B. 30√3 m (approx. 51.96 m)
3. C. 60 m
4. D. 15√3 m

Show the answer

Answer: B. 30√3 m (approx. 51.96 m). Angle of elevation from boat to cliff top = angle of depression = 30° (alternate angles). tan 30° = opposite / adjacent = 30 / d → 1/√3 = 30 / d → d = 30√3 metres.

A model question written by Idasara on the exam pattern. It is not from a past paper.

Matrices

What you must know

• Order compatibility for operations: For addition/subtraction, orders must be EXACTLY identical; for multiplication (A × B), inner dimensions must match: (2 × 3) can multiply (3 × 2), but (2 × 3) CANNOT multiply (2 × 3).
• Matrix multiplication is non-commutative: In general, AB ≠ BA; never assume matrix products can be reversed.
• Row-by-column multiplication arithmetic: Entry c₁₁ is (row 1 of A) dot (col 1 of B): a₁₁b₁₁ + a₁₂b₂₁; take extreme care with negative signs.

Try one question

Given matrix A = [[2, -1], [3, 4]] and matrix B = [[1, 5], [0, -2]], find the matrix A + 2B.

1. A. [[4, 9], [3, 0]]
2. B. [[3, 4], [3, 2]]
3. C. [[2, 9], [3, 0]]
4. D. [[4, 8], [3, 2]]

Show the answer

Answer: A. [[4, 9], [3, 0]]. Scalar multiplication: 2B = [[2, 10], [0, -4]]. Matrix addition: A + 2B = [[2 + 2, -1 + 10], [3 + 0, 4 + (-4)]] = [[4, 9], [3, 0]].

A model question written by Idasara on the exam pattern. It is not from a past paper.

Inequalities

What you must know

• Negative sign reversal trap: Forgetting to reverse inequality sign when dividing by negative coefficient is the single most common mark-losing trap.
• Integral Solutions: Pay close attention to domain restrictions: 'positive integers' (x ∈ {1, 2, 3, ...}), 'whole numbers' (x ∈ {0, 1, 2, ...}), or 'integers' (x ∈ {..., -1, 0, 1, ...}).
• Compound Inequalities: For a < bx + c ≤ d, solve the two simultaneous inequalities independently or apply inverse operations simultaneously across all three parts.

Try one question

Solve the compound linear inequality: -3 ≤ 2x + 1 < 9, where x ∈ ℝ.

1. A. -1 ≤ x < 5
2. B. -4 ≤ x < 8
3. C. -2 < x ≤ 4
4. D. -2 ≤ x < 4

Show the answer

Answer: D. -2 ≤ x < 4. Subtract 1 from all parts: -3 - 1 ≤ 2x < 9 - 1 → -4 ≤ 2x < 8. Divide by 2: -2 ≤ x < 4.

A model question written by Idasara on the exam pattern. It is not from a past paper.

Cyclic Quadrilaterals

What you must know

• Tangent-Chord (Alternate Segment) interplay: In Paper 2 riders, cyclic quadrilateral angle theorems are frequently combined with the alternate segment theorem (angle between tangent and chord equals angle in alternate segment).
• Opposite vs Adjacent angles: Only OPPOSITE angles add to 180°; adjacent angles generally DO NOT add to 180° unless the cyclic quadrilateral is also an isosceles trapezium or rectangle.
• Angle in a Semicircle: If a cyclic quadrilateral diagonal passes through circle center (diameter), the angle subtended on the circumference is 90°.

Try one question

In a cyclic quadrilateral ABCD, the opposite angles satisfy angle A = 3x and angle C = 2x + 20°. Find the magnitude of angle A.

1. A. 96°
2. B. 100°
3. C. 32°
4. D. 84°

Show the answer

Answer: A. 96°. Opposite angles of a cyclic quadrilateral are supplementary: angle A + angle C = 180° → 3x + (2x + 20°) = 180° → 5x = 160° → x = 32°. Therefore, angle A = 3(32°) = 96°.

A model question written by Idasara on the exam pattern. It is not from a past paper.

Tangents

What you must know

• Alternate segment chord tracing: Identify the chord meeting the tangent at point of contact; trace that chord to the circumference opposite — THAT is the equal alternate segment angle.
• External tangents right triangle: The line joining the external point to the centre (OP) bisects the angle between the tangents and forms two congruent right-angled triangles with the radii.
• Geometric reason notation: In Paper 2 proofs, cite 'tangent ⊥ radius', 'tangents from ext. point =', or 'alternate segment theorem'.

Try one question

PT is a tangent drawn from an external point P to a circle at point of contact T. If O is the centre of the circle, radius OT = 5 cm, and OP = 13 cm, find the length of the tangent PT.

1. A. 10 cm
2. B. 14 cm
3. C. 8 cm
4. D. 12 cm

Show the answer

Answer: D. 12 cm. By theorem, the tangent at any point of a circle is perpendicular to the radius through the point of contact (angle OTP = 90°). In right triangle OTP: PT² + OT² = OP² → PT² + 5² = 13² → PT² + 25 = 169 → PT² = 144 → PT = 12 cm.

A model question written by Idasara on the exam pattern. It is not from a past paper.

Constructions

Sets

Probability