Step-by-step worked examples and graded practice questions on trigonometric ratios — the sine rule (including the ambiguous case), the cosine rule, finding the area of a triangle, exact trig values, and the graphs of sine and cosine. Written to the Edexcel Pure Year 1 specification and equally suitable for AQA and OCR A.
📚 Pure Year 1 (AS)✅ 15 Practice Questions🔍 6 Worked Examples⚠️ Common Mistakes
Struggling with trigonometry?
Alamin diagnoses exactly which skills are missing and builds a structured plan to fix them — backed by AI-powered practice between sessions.
GCSE trigonometry was built entirely around right-angled triangles. A-Level extends this to any triangle, using the sine rule and cosine rule, and separately looks at trig functions as full graphs rather than single-triangle ratios.
Four skills make up this topic:
The sine rule — relating sides and angles in any triangle, including the ambiguous case
The cosine rule — finding a side or angle when the sine rule doesn't directly apply
Area of a triangle — using two sides and the included angle, without needing the height
Graphs of sine and cosine — reading off exact values and solving equations graphically
The sine rule
In any triangle \(ABC\), with side \(a\) opposite angle \(A\), side \(b\) opposite angle \(B\), and side \(c\) opposite angle \(C\):
Triangle \(ABC\) from Worked Example 1, drawn to scale from the given angles — side \(a\) is opposite angle \(A\); side \(b\) (amber) is the unknown being solved for.
When you're given two sides and a non-included angle (SSA), there can be two valid triangles — this is the ambiguous case. It arises when the given angle is acute and the side opposite it is shorter than the other given side.
Worked Example 2
In triangle \(ABC\), \(a = 7\) cm, \(b = 10\) cm, and angle \(A = 30°\). Find the two possible values of angle \(B\).
1
Use \(\dfrac{\sin B}{b} = \dfrac{\sin A}{a}\): \(\sin B = \dfrac{10 \sin 30°}{7} = \dfrac{5}{7}\)
2
Principal value: \(B_1 = \sin^{-1}\left(\dfrac{5}{7}\right) = 45.6°\)
In triangle \(ABC\), \(a = 8\) cm, \(b = 6\) cm, \(c = 5\) cm. Find angle \(A\).
1
Substitute into \(\cos A = \dfrac{b^2+c^2-a^2}{2bc}\): \(\cos A = \dfrac{36+25-64}{2(6)(5)}\)
2
Evaluate: \(\cos A = \dfrac{-3}{60} = -0.05\)
3
Since \(\cos A\) is negative, \(A\) is obtuse: \(A = \cos^{-1}(-0.05)\)
Answer\(A = 92.9°\) (1 d.p.)
Area of a triangle
Given two sides and the included angle, there's no need to find the height first:
\[\text{Area} = \frac{1}{2}ab\sin C\]
Worked Example 5
Find the area of triangle \(ABC\) where \(b = 8\) cm, \(c = 10\) cm, and angle \(A = 50°\).
1
Substitute into \(\text{Area} = \dfrac{1}{2}bc\sin A\): \(\text{Area} = \dfrac{1}{2}(8)(10)\sin 50°\)
2
Evaluate: \(\text{Area} = 40 \times 0.766\)
AnswerArea \(= 30.6\) cm² (3 s.f.)
Exact trigonometric values
For angles \(0°, 30°, 45°, 60°\) and \(90°\), sine, cosine and tangent take exact values — memorise these rather than relying on a calculator, since exam questions often require an exact (surd) answer.
\(0°\)
\(30°\)
\(45°\)
\(60°\)
\(90°\)
\(\sin\)
\(0\)
\(\dfrac{1}{2}\)
\(\dfrac{\sqrt{2}}{2}\)
\(\dfrac{\sqrt{3}}{2}\)
\(1\)
\(\cos\)
\(1\)
\(\dfrac{\sqrt{3}}{2}\)
\(\dfrac{\sqrt{2}}{2}\)
\(\dfrac{1}{2}\)
\(0\)
\(\tan\)
\(0\)
\(\dfrac{\sqrt{3}}{3}\)
\(1\)
\(\sqrt{3}\)
undefined
Graphs of sine and cosine
The graphs of \(y=\sin\theta°\) and \(y=\cos\theta°\) both repeat every \(360°\) (they're periodic), oscillate between \(-1\) and \(1\), and are simply horizontal translations of each other. Reading across a horizontal line on the graph shows every solution to a trig equation in one go — including the second solution that's easy to miss algebraically.
\(y=\sin\theta°\) (teal) and \(y=\cos\theta°\) (amber) for \(0° \leq \theta \leq 360°\). The dashed line at \(y=0.5\) crosses the sine curve at \(\theta=30°\) and \(\theta=150°\) — the two solutions found in Worked Example 6.
Worked Example 6
Use the graph of \(y=\sin\theta°\) to solve \(\sin\theta = 0.5\) for \(0° \leq \theta \leq 360°\).
1
Find the principal value: \(\theta = \sin^{-1}(0.5) = 30°\)
2
The sine graph is symmetrical about \(\theta=90°\), so the second solution is \(180° - 30° = 150°\)
3
No other crossings of \(y=0.5\) occur in \(0°\) to \(360°\) — sine is negative for the rest of the range past \(180°\)
Answer\(\theta = 30°\) or \(\theta = 150°\)
Exam tips
💡 Tip 1
Check which rule actually applies before starting
Sine rule needs a complete angle–side pair; cosine rule is for three sides, or two sides and the included angle. Picking the wrong rule is the most common wasted-time mistake in this topic.
💡 Tip 2
Always check for the ambiguous case in SSA problems
If you're given two sides and a non-included angle, ask whether \(180°\) minus your answer also gives a valid triangle (angles summing to less than \(180°\)) before assuming there's only one solution.
💡 Tip 3
A negative cosine means an obtuse angle — that's expected, not an error
If \(\cos A\) comes out negative, don't assume you've made a mistake — it correctly signals that \(A\) is between \(90°\) and \(180°\).
💡 Tip 4
Label which angle is "included" carefully in the area formula
\(\text{Area} = \frac{1}{2}ab\sin C\) only works when \(C\) is the angle between the two sides \(a\) and \(b\) — using a different angle gives the wrong area entirely.
💡 Tip 5
Use the graph's symmetry, not just your calculator, to find every solution
A calculator only ever returns the principal value — sketching (or picturing) the graph is what reveals the second solution within the given range.
Common mistakes
Common Mistake 1
Mismatching sides and angles in the sine rule
Side \(a\) must be paired with angle \(A\) (the angle opposite it), never with a different angle — double-check which side is opposite which angle before substituting.
Common Mistake 2
Forgetting the ambiguous case entirely
Giving only the calculator's principal value when a question genuinely has two valid triangles is one of the most common lost marks in sine rule questions — always check whether \(180°\) minus your answer is also valid.
Common Mistake 3
Taking the square root too early in the cosine rule
Calculate \(a^2\) fully first, then square root at the very end — square-rooting partway through (e.g. before subtracting \(2bc\cos A\)) gives a meaningless result.
Common Mistake 4
Rounding intermediate values
Round only your final answer — carrying a rounded \(\cos A\) or \(\sin B\) into the next step compounds the error and can shift the final answer outside the accepted range.
Common Mistake 5
Using a decimal approximation where an exact value is required
If a question says "give your answer as a surd" or "in exact form", writing \(0.866\) instead of \(\dfrac{\sqrt{3}}{2}\) loses marks even if the decimal is numerically correct.
Practice questions
Work through each question before checking the worked solution.
Core Skills
Q1In triangle \(ABC\), angle \(A=50°\), angle \(B=70°\), and \(a=6\) cm. Find side \(b\).Core Skills
Q2In triangle \(ABC\), \(b=12\) cm, \(c=9\) cm, angle \(A=60°\). Find side \(a\).Core Skills
Q3Find the area of triangle \(ABC\) where \(b=7\) cm, \(c=5\) cm, angle \(A=80°\).Core Skills
Q4Write down the exact value of \(\sin 60°\).Core Skills
Q5State the period of the graph \(y=\sin\theta°\).Core Skills
Exam-Style
Q6In triangle \(ABC\), \(a=9\) cm, \(b=11\) cm, angle \(A=40°\). Find the two possible values of angle \(B\).Exam-Style
Q7In triangle \(ABC\), \(a=10\) cm, \(b=8\) cm, \(c=7\) cm. Find the largest angle.Exam-Style
Q8Find the area of triangle \(ABC\) given \(a=6\) cm, \(b=9\) cm, angle \(C=45°\).Exam-Style
Q9Solve \(\cos\theta = 0.5\) for \(0° \leq \theta \leq 360°\).Exam-Style
Q10Write down the exact value of \(\tan 45°\).Exam-Style
A* Challenge
Q11In triangle \(ABC\), angle \(B=35°\), angle \(C=75°\), and side \(a=15\) cm. Find side \(b\).A* Challenge
Q12The area of triangle \(ABC\) is \(24\) cm², with \(a=8\) cm and \(b=10\) cm. Find the two possible values of angle \(C\).A* Challenge
Q13In triangle \(ABC\), \(AB=13\) cm, \(BC=14\) cm, \(CA=15\) cm. Find the size of the largest angle, to 1 decimal place.A* Challenge
Q14In triangle \(ABC\), angle \(A=112°\), \(b=6\) cm, \(c=9\) cm. Find the area of the triangle, to 3 significant figures.A* Challenge
\(24 = \dfrac{1}{2}(8)(10)\sin C \implies \sin C = 0.6\)
2
\(C_1 = \sin^{-1}(0.6) = 36.9°\)
\(C = 36.9°\) or \(180°-36.9°=143.1°\)
Q13 — \(67.4°\)
1
Longest side is \(CA=15\), opposite angle \(B\). With \(a=BC=14\), \(c=AB=13\): \(\cos B = \dfrac{14^2+13^2-15^2}{2(14)(13)}\)
2
\(\cos B = \dfrac{140}{364} = 0.385\)
\(B = 67.4°\) (1 d.p.)
Q14 — Area \(= 25.0\) cm²
1
Area \(= \dfrac{1}{2}(6)(9)\sin 112° = 27\sin 112°\)
2
\(\sin 112° = \sin 68° = 0.9272\)
Area \(= 25.0\) cm² (3 s.f.)
Q15 — \(\theta = 210°\) or \(330°\)
1
Sine is negative for \(180° < \theta < 360°\); reference angle \(\sin^{-1}(0.5)=30°\)
2
Solutions: \(180°+30°=210°\) and \(360°-30°=330°\)
\(\theta = 210°\) or \(330°\)
Summary
Use the sine rule, \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\), whenever you have a complete angle–side pair — and always check for the ambiguous case in SSA problems.
Use the cosine rule, \(a^2=b^2+c^2-2bc\cos A\), for three sides or two sides plus the included angle. A negative cosine correctly signals an obtuse angle.
Find a triangle's area from two sides and the included angle with \(\text{Area}=\frac{1}{2}ab\sin C\) — no height needed.
Memorise the exact trig values at \(0°, 30°, 45°, 60°, 90°\) for surd-form answers.
The graphs of \(y=\sin\theta°\) and \(y=\cos\theta°\) are periodic (period \(360°\)) and symmetric — use that symmetry to find every solution to a trig equation, not just the calculator's principal value.
If trigonometric ratios are still causing problems, Alamin's diagnostic approach identifies exactly which skills are missing and builds a targeted plan to address them — with AI-powered practice between sessions.