Sheet № 111 · Higher only · AQA · Edexcel · OCR
Sine Rule –
The sine rule lets you work with non-right-angled triangles when you have a matching angle-side pair. It is one of the two major Higher-tier trigonometry rules and appears regularly on AQA, Edexcel and OCR papers.
§Key definitions
Question:
In triangle ABC, angle A = 50°, angle B = 70° and side a = 10 cm. Find side b to 1 decimal place.
Answer:
b = 12.3 cm (1 d.p.).
Q1 (Foundation):
In triangle ABC, angle A = 45°, angle C = 65° and side a = 8 cm. Find side c to 1 d.p.
Q2 (Foundation):
In triangle DEF, angle D = 55°, side d = 12 cm and side e = 10 cm. Find angle E to 1 d.p.
Q3 (Higher):
In triangle XYZ, angle X = 38°, angle Y = 72° and side y = 25 cm. Find side x to 1 d.p.
§Formulas to memorise
a/sin A = b/sin B = c/sin C (for finding sides)
sin A/a = sin B/b = sin C/c (for finding angles)
For a side: use a/sin A = b/sin B. Substitute and solve for the unknown side.
For an angle: use sin A/a = sin B/b. Substitute, solve for sin of the unknown angle, then use sin⁻¹.
Worked example
In triangle ABC, angle A = 50°, angle B = 70° and side a = 10 cm. Find side b to 1 decimal place.
Working: Use the sine rule for finding a side: a/sin A = b/sin B 10/sin 50° = b/sin 70° 10/0.7660 = b/0.9397 13.055 = b/0.9397 b = 13.055 × 0.9397
⚠ Common mistakes
- ✗Mismatching sides and angles. Side a must be opposite angle A. If you pair a side with the wrong angle, the answer is wrong.
- ✗Forgetting to use sin⁻¹ when finding an angle. If sin Q = 0.6331, the angle is sin⁻¹(0.6331), not 0.6331.
- ✗Using the sine rule when the cosine rule is needed. The sine rule requires a complete pair. If you have SAS (two sides and the included angle) or SSS (three sides), you need the cosine rule instead.
✦ Exam tips
- →Copy the formula from the formula sheet and write it into your working — this earns a method mark.
- →Always check: do I have a complete angle-side pair? If yes, use the sine rule. If no, consider the cosine rule.
- →For the ambiguous case, if sin⁻¹ gives you an acute angle but the diagram suggests an obtuse angle, the second solution is 180° minus the acute angle.