AQA A-level Mathematics 7357, section E (Trigonometry), version 1.3 (first assessment 2018). Original teaching notes; objective-level mapping lives in content/teaching/aqa_a_level_mathematics/objective-review.yaml.
Prerequisites: GCSE right-triangle trigonometry and exact values. Each lesson develops its own method before independent transfer.
These are original teaching examples, not official AQA material.
Two students are given A = 30°, a = 7 and b = 10. One reports one triangle; the other insists there are two, and both are right. What information separates the cases?
Start with a prediction. State which rule pairs sides with opposite angles, and what extra check an inverse-sine answer needs.
Build the model
On the unit circle 单位圆, the point at angle θ is (cos θ, sin θ), so signs follow the quadrant and exact values come from the 30-60-90 and 45-45-90 triangles. A radian 弧度 is the angle whose arc equals the radius: s = rθ and sector area = r²θ/2 require radians.
The sine rule 正弦定理 pairs opposite sides and angles: a/sin A = b/sin B. The cosine rule 余弦定理 a² = b² + c² − 2bc cos A uses the angle opposite a. In the ambiguous case 二义情形, sin B = b sin A / a can give B or its supplement 180° − B — both must be tested against the remaining angle.
unit circle: The circle of radius 1 on which (cos θ, sin θ) is the point at angle θ; radian: The angle subtended by an arc of length equal to the radius; sine rule: The rule a/sin A = b/sin B pairing opposite sides and angles; cosine rule: The rule a² = b² + c² − 2bc cos A; ambiguous case: The sine-rule situation where two triangles may fit the same data.
Choose evidence that can test it
Choose the rule from the known information: a matching side-angle pair calls for the sine rule; two sides with the INCLUDED angle, or three sides, call for the cosine rule. For area, ab sin C / 2 needs the angle BETWEEN a and b.
Test each candidate angle: the remaining angle C = 180° − A − B must be positive, and every side positive. A calculator's inverse sine returns only the acute principal value — the supplement is a separate candidate.
Work from known quantities
State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
Known: sides 6 and 8 enclose 60°. Use the cosine rule: c² = 6² + 8² − 2 × 6 × 8 × cos 60° = 36 + 64 − 96 × 0.5 = 52. So c = √52 = 2√13 ≈ 7.21. The area is (1/2) × 6 × 8 × sin 60° = 24 × (√3/2) = 12√3.
Example:
For A = 30°, a = 7, b = 10, find the possible values of B using the same sequence: known → sine rule → both branches → validity check. (Answer: sin B = 10 × sin 30°/7 = 5/7, so B ≈ 45.6° or B ≈ 180° − 45.6° = 134.4°. Both leave a positive third angle, so TWO triangles fit.)
Check the conclusion and its limits
The ambiguous case needs the count: with h = b sin A and acute A, a < h gives none, a = h one right triangle, h < a < b gives two, and a ≥ b gives one. Trig equations on an interval must list every solution: sin x = 1/2 on [0, 2π) has both π/6 and 5π/6.
Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Warn:
The calculator's inverse-sine value is the only possible angle in a sine-rule problem. This claim is false: sine is positive in two quadrants, so the supplement 180° − B is a second candidate and must be checked, which is exactly the ambiguous case.
Key:
Trigonometry: unit circle for values and signs, radians for arcs and sectors, sine rule with its ambiguous case, cosine rule for included angles, and ab sin C / 2 for area — always with a validity check on the remaining angle.
