The Sine Rule, the Cosine Rule, and the Ambiguous Case

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

Four shapes of triangle information, and which rule actually fits each one

Spec 3.1 gives three tools for a general triangle — one whose angles are not all known, and not a right angle in sight: the sine rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ (lower-case side opposite the matching upper-case angle, throughout); the cosine rule, $a^2 = b^2 + c^2 - 2bc\cos A$ , and its rearrangement for an angle, $\cos A = \frac{b^2+c^2-a^2}{2bc}$ ; and the area formula, $\text{Area} = \frac12 ab\sin C$ , where $C$ is the angle trapped BETWEEN the two sides $a$ and $b$ being multiplied. None of the three is a general-purpose tool — each is built for a specific shape of given information, and recognising the shape is the actual first step of every question on this spec point, before any arithmetic starts.

SSS (three sides, no angle): only the cosine rule can start here, rearranged for an angle — $\cos A = \frac{b^2+c^2-a^2}{2bc}$ . The sine rule needs a full angle-and-opposite-side pair to build its ratio, and SSS data contains none.

SAS (two sides and the angle trapped between them): only the cosine rule can start here too, this time in its original form, $a^2 = b^2+c^2-2bc\cos A$ — the unknown side is opposite the KNOWN angle, and the two known sides are the ones enclosing it. Once that one new side is found, the triangle has a full opposite pair and the sine rule can take over for anything left.

ASA or AAS (two angles, and any one side): the sine rule handles this cleanly and — this matters — without any ambiguity at all. Once two angles are known, the third is fixed by the angle sum ( $180°$ minus the other two), and finding a missing SIDE from a full opposite pair is a single division, never a $\sin^{-1}$ . There is nothing here for the ambiguous case to attach to; it is coming later, and it needs a genuinely different data shape to appear.

SSA (two sides and a non-included angle — an angle that is NOT trapped between the two given sides): the sine rule is still the only tool that fits, but this is the one shape where finding the missing angle means taking $\sin^{-1}$ of a ratio — and that is precisely where the ambiguous case lives. Nothing about SSS, SAS or ASA/AAS ever raises this question; it is specific to this one combination, which is exactly why spec 3.1 names it by name rather than leaving it as a general warning about the sine rule.

Where the area formula comes from, and what the exam formula booklet actually gives you

The area formula follows in one line from a fact already familiar from GCSE: area of a triangle is $\frac12 \times \text{base} \times \text{height}$ . Take sides $a$ and $b$ meeting at angle $C$ , and use $a$ as the base. Drop a perpendicular from the vertex opposite $a$ down to the line containing $a$ ; in the right-angled triangle this perpendicular creates, the hypotenuse is $b$ and the angle at the base is $C$ , so $\text{height} = b\sin C$ (opposite over hypotenuse). Substitute: $\text{Area} = \frac12 \times a \times b\sin C = \frac12 ab\sin C$ . Nothing about this needs $a$ and $b$ to be any particular pair of sides — it needs $C$ to be the angle physically BETWEEN whichever two sides are being multiplied, because that is the one fact the derivation actually used.

The exam formula booklet is genuinely thin here, and knowing exactly how thin changes how this topic should be revised. Verified directly from the booklet's "Pure Mathematics P1" section: it contains exactly two entries for the whole of P1 — mensuration (sphere surface area, cone curved surface) and the cosine rule, $a^2 = b^2+c^2-2bc\cos A$ . That is the entire P1 section. The sine rule, the area formula, and both spec-3.2 formulae (arc length, sector area) are NOT there — the specification states plainly that formulae students are expected to know "are given below and will not appear in the booklet." So on this exact spec point: one of the three tools (cosine rule) is provided on the day, and two of the three (sine rule, area formula) have to come from memory. Losing marks to "I knew which rule to use but couldn't recall it exactly" is a genuinely different failure from an algebra mistake, and it is worth treating as its own thing to practise — not folded silently into "revise trigonometry."

Worked, in full

Sine rule for a missing side (ASA) — no ambiguity, because both angles are given directly

01
In triangle $ABC$ , angle $A = 42°$ , angle $B = 71°$ , and side $a = 8.5$ cm (opposite $A$ ). Find side $b$ (opposite $B$ ). Before any calculation: both angles are GIVEN, not derived by $\sin^{-1}$ , so there is no fork to worry about here — this is the safe, unambiguous half of the sine rule, worth doing once cleanly before the risky half arrives.
Earns: No mark of its own — recognising the data shape (ASA, not SSA) is what tells you this question needs no ambiguous-case check at all.
02
Set up the sine rule with the two known opposite pairs: $\frac{a}{\sin A} = \frac{b}{\sin B}$ , so $\frac{8.5}{\sin 42°} = \frac{b}{\sin 71°}$ .
Earns: M1 — a correct attempt at the sine rule with sides and angles correctly paired. Quoting it in this general form BEFORE substituting is the same safety net the general marking guidance names for the quadratic formula: the method mark can be read off directly here, rather than having to be inferred from correct-looking numbers later.
03
Rearrange and evaluate: $b = \frac{8.5\sin 71°}{\sin 42°} = \frac{8.5 \times 0.9455}{0.6691} \approx 12.0$ cm (3 s.f.).
Earns: A1 — correct value of $b$ . The two sines are close in size here ( $0.9455$ vs $0.6691$ ), so $b$ comes out noticeably larger than $a$ — worth a half-second sanity check: the side opposite the BIGGER angle should always be the LONGER side, and $71° > 42°$ , so $b > a$ is exactly what should happen.
04
The third angle and side follow the same safe pattern. $C = 180° - 42° - 71° = 67°$ , and $c = \frac{8.5\sin 67°}{\sin 42°} \approx \frac{8.5 \times 0.9205}{0.6691} \approx 11.7$ cm (3 s.f.).
Earns: A1 — correct value of $c$ , using the SAME denominator as before ( $\sin 42°$ , from the one full opposite pair the question actually gave). A common slip here is re-deriving $\sin A$ from a rounded intermediate value instead of the original given angle — always divide by the sine of the angle whose opposite side was given in the question, not a side found partway through.

Mechanism

Where the second triangle actually comes from — a compass, not just an equation

Take the genuinely ambiguous shape: angle $A$ , the side $a$ opposite it, and one other side $b$ — with $A$ acute and $b > a$ . Place vertex $A$ at the origin, with side $AB$ running along the positive x-axis (wherever $B$ turns out to be), and place $C$ at distance $b$ from $A$ , at angle $A$ above that axis: $C = (b\cos A,\ b\sin A)$ . Vertex $B$ has to be SOMEWHERE on that positive x-axis, and it has to satisfy exactly one more condition: $BC = a$ . The set of every point at distance $a$ from $C$ is a circle of radius $a$ centred on $C$ — so $B$ is wherever that circle crosses the x-axis, and a circle can cross a line at 0, 1, or 2 points. Concretely, with $A=30°$ , $a=5$ , $b=9$ : $C = (9\cos 30°,\, 9\sin 30°) \approx (7.79,\, 4.5)$ , and solving $(t - 7.79)^2 + 4.5^2 = 5^2$ for the x-axis crossing points $t$ gives $t \approx 7.79 \pm \sqrt{5^2 - 4.5^2} \approx 7.79 \pm 2.18$ — TWO values, $t \approx 9.97$ and $t \approx 5.61$ , both positive, both genuine positions for $B$ . Checking the angle at $B$ for each (the angle between $BA$ and $BC$ ) confirms the two triangles are genuinely different shapes, not the same one measured twice: the FARTHER crossing point ( $t \approx 9.97$ , further from $A$ ) gives $B \approx 64.2°$ , an acute angle; the NEARER crossing point ( $t \approx 5.61$ , closer to $A$ ) gives $B \approx 115.8°$ , obtuse. Both are legitimate triangles built from the exact same $A$ , $a$ and $b$ — the algebra's two solutions aren't a glitch, they are two real, differently-shaped triangles the given data genuinely doesn't distinguish between on its own.

Diagram — Same A, a and b — two different triangles, drawn from where the circle of radius a meets the baseline

x-axis: position of B along the ray from A · y-axis: height above that ray

Ray from A carrying vertex B: Fixed once angle A and the direction of AB are chosen. Vertex B must sit somewhere on this ray — the only question is where.
Side AC, length b, at angle A to the ray: Fixed by the given angle A and given length b. This segment does not change between the two triangles — only where B lands on the ray does.
Circle of radius a, centred at C: Every point exactly distance a from C — the set of every place BC could possibly end, since side BC is fixed at length a. Where this circle crosses the ray from A is where B can be.

B₁ — the far crossing point: Farther from A along the ray. Gives the ACUTE version of angle B. This is the triangle a calculator's sin⁻¹ key returns by default.
B₂ — the near crossing point: Closer to A along the ray, still on the positive side (still a valid triangle vertex, not behind A). Gives the OBTUSE version of angle B — the one sin⁻¹ never mentions on its own.
C — fixed throughout: The one point that does not move between the two triangles. Both B₁ and C, and B₂ and C, are joined by a side of the same length a — that is the whole reason both crossing points count as valid answers to the same question.

Common error: Solving the sine rule for the missing angle, taking whatever a calculator's sin⁻¹ key returns, and treating that single acute value as the finished answer — with no check for whether a second, obtuse-angled triangle is equally valid.

Correct: Every time the sine rule is used to find an angle from an SSA (two-sides-and-a-non-included-angle) setup, check explicitly for the supplementary angle, 180° minus the calculator's value, and decide — from the question's own wording, an extra given length, or the angle sum simply failing to fit — which one, or both, the triangle actually requires.

examiner-report · Jan 2019 · Q7

Mechanism

Why the cosine rule can never do this to you — and why the sine rule can

This isn't a difference in how carefully each rule is usually applied; it's a structural difference in the two functions themselves, over the only range a triangle's angles can actually occupy: $0° < \theta < 180°$ . Over that range, $\sin\theta$ rises from 0 up to 1 at $\theta = 90°$ , then falls back down to 0 — a single hump, symmetric about $90°$ , so that $\sin(180° - \theta) = \sin\theta$ for every $\theta$ in the range. Every value of $\sin\theta$ strictly between 0 and 1 is therefore in $(0°, 180°)$ : taking $\sin^{-1}$ of a ratio can never, on its own, tell you which one a real triangle actually has. Now run the same check on $\cos\theta$ over the identical range: it starts at $\cos 0° = 1$ and falls the entire way to $\cos 180° = -1$ , strictly decreasing throughout (its gradient is $-\sin\theta$ , and $\sin\theta$ is strictly positive everywhere on the open interval $(0°,180°)$ , so the fall never pauses or reverses). A function that is strictly decreasing over its whole domain — so $\cos^{-1}$ of any number between $-1$ and $1$ returns exactly one angle in $(0°, 180°)$ , with nothing left over to check. This is also why a negative value partway through a cosine-rule calculation is not a warning sign: $\cos\theta$ is negative for every obtuse angle in the valid range, so an obtuse result simply looks like a negative number under $\cos^{-1}$ — a completely ordinary, correct outcome, not a mistake to hunt down.

Worked, in full

Cosine rule for a missing side (SAS)

01
In triangle $ABC$ , $b = 6$ cm, $c = 9$ cm, and the INCLUDED angle $A = 50°$ (the angle trapped between the two given sides). Find side $a$ . Recognise the shape first: two sides and the angle between them is exactly what the cosine rule's original form is built for — the sine rule cannot even start here, since neither given side yet has a known opposite angle.
Earns: No mark of its own — confirming the SAS shape is what licenses using the cosine rule's side-finding form at all.
02
Quote the cosine rule with the correct side matched to the correct angle: $a^2 = b^2 + c^2 - 2bc\cos A$ , with $A$ specifically the angle BETWEEN the two sides being substituted.
Earns: M1 — correct method: the general marking guidance's advice to quote a memorised formula before substituting applies here exactly as it does to the quadratic formula, and is the safety net against losing this mark to a substitution slip.
03
$a^2 = 6^2 + 9^2 - 2(6)(9)\cos 50° = 36 + 81 - 108(0.6428) \approx 117 - 69.42 \approx 47.58$ .
Earns: A1 — correct value inside the square root. The $-2bc\cos A$ term is where a sign error costs the whole answer: $\cos 50°$ is positive (acute angle), so this term SUBTRACTS from $b^2+c^2$ here — for an obtuse included angle it would instead ADD, since cosine turns negative past 90°.
04
$a = \sqrt{47.58} \approx 6.90$ cm (3 s.f.).
Earns: A1 — correct final value. Only the positive square root is taken: a side length can't be negative, so unlike the sine rule's angle-finding case, there is no second candidate to weigh up here at all — a length has exactly one valid sign, full stop.

Complete it yourself

Complete the chain — cosine rule for a missing angle (SSS), from three sides and no calculator shortcut

01
Triangle $ABC$ has $a = 7$ cm, $b = 8$ cm, $c = 13$ cm. Find angle $A$ (opposite the shortest side). SSS data — the sine rule has no opposite pair to start from, so the cosine rule, rearranged for an angle, is the only tool that fits.
02
Rearrange the cosine rule for the angle: $\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{8^2+13^2-7^2}{2(8)(13)} = \frac{64+169-49}{208} = \frac{184}{208} \approx 0.8846$ .

Worked, in full

Area of a triangle — the one formula where using the wrong angle produces a wrong but plausible-looking number

01
Triangle $ABC$ has $a = 8$ cm, $b = 11$ cm, and the angle BETWEEN them, $C = 63°$ . Find the area. Confirm first which angle is actually trapped between the two sides being used — the formula $\frac12 ab\sin C$ only works with $C$ as the INCLUDED angle, never any other angle in the triangle, however many are given.
Earns: No mark of its own — identifying which angle is included is what the whole calculation depends on.
02
$\text{Area} = \frac12 ab\sin C = \frac12 (8)(11)\sin 63° = 44\sin 63° \approx 44 \times 0.8910 \approx 39.2$ cm² (3 s.f.).
Earns: M1 for a correct attempt at $\frac12 ab\sin C$ with the included angle correctly identified, A1 for the correct final value. Losing the $\frac12$ here doubles the answer to $\approx 78.4$ — a wrong number that still looks like a plausible area, which is exactly why it is worth checking for on sight rather than trusting that an implausible-looking result would always give itself away.

In your own words

In one sentence: why can solving a triangle's angle with the sine rule sometimes need a check that solving for a side with the sine rule, or solving for anything at all with the cosine rule, never needs?

Marked, line by line

In triangle $ABC$ , $AB = 10$ cm, $BC = 7$ cm, and angle $BAC = 35°$ . Given that angle $ACB$ is obtuse: (a) find angle $ACB$ , (3) (b) hence find angle $ABC$ , (1) (c) hence find the area of triangle $ABC$ . (3) — VERIDIAN-original question, built directly around the one verified WMA11 finding on this spec point (Jan 2019, Q7: "The majority used the sine rule and achieved the correct acute angle. However, many failed to find the required obtuse angle."). Not a reproduction of that question — the facts bank preserves only the finding, not Q7's own numbers or wording, so every length, angle and mark allocation below is VERIDIAN-original, modelled on verified mark-scheme conventions rather than transcribed from a real scheme, which for an original question does not exist.

7 marks available

(a) — 3 marks

01
$\frac{\sin C}{AB} = \frac{\sin A}{BC}$ , i.e. $\frac{\sin C}{10} = \frac{\sin 35°}{7}$ (side $AB=c=10$ is opposite angle $C$ ; side $BC=a=7$ is opposite the given angle $A=35°$ ).
Method mark for the sine rule with sides and angles correctly paired. Quoting it in this form, before substituting, is the safety net the general marking guidance names — the method mark is readable directly here rather than needing to be inferred later.
M1
02
$\sin C = \frac{10\sin 35°}{7} \approx 0.8194$ , so $C \approx 55.0°$ or $C \approx 180° - 55.0° = 125.0°$ .
Accuracy mark, dependent on the M1, for reaching a valid candidate angle from $\sin^{-1}$ of the ratio. The real WMA11 mark scheme for this exact ambiguous-case question type (Jan 2019, Q7(a), verified against the primary-source PDF) awards this mark for EITHER acceptable candidate on its own — its own wording is "∠ACB awrt(52 or 53)° or awrt(127 or 128)°" — so this is genuinely earned the moment $C \approx 55.0°$ appears, before the supplementary angle is even found. It is the NEXT mark, not this one, that the examiner report's finding is actually about.
A1
03
The question states angle $ACB$ is obtuse, so $C = 125.0°$ (1 d.p.), not the acute candidate.
Accuracy mark, correct answer only (cao) — the real mark scheme's equivalent line for this question type reads "∠ACB = 127.5° only", meaning no other value earns it, whatever candidate values were shown upstream. This is the mark a correct-but-incomplete script loses: reaching either candidate value already banks the mark above, but the working still has to end by choosing the one the geometry demands, not the one a calculator returns first.
A1

(b) — 1 mark

101
$B = 180° - A - C = 180° - 35° - 125.0° = 20.0°$ .
Independent mark, follow through their C from part (a). The angle-sum method is correct whichever C a candidate is carrying forward — this mark checks the arithmetic of 180 minus the other two, not which branch of part (a) was chosen.
B1 ft

(c) — 3 marks

201
Area $= \frac12 \times AB \times BC \times \sin B$ (AB and BC are the two sides meeting at vertex B, so angle B is the angle physically INCLUDED between them).
Method mark for a correct attempt at the area formula with the genuinely included angle identified — AB and BC share vertex B, so B, not A or C, is the angle the formula needs here.
M1
202
Area $= \frac12 (10)(7)\sin 20.0° = 35\sin 20.0° \approx 35 \times 0.3424 \approx 11.98$ .
Accuracy mark for correct substitution and value, dependent on a correct B from part (b).
A1
203
Area $\approx 12.0$ cm² (3 s.f.).
Correct answer only, against the TRUE value built on the obtuse angle the question actually specified — not follow-through-able against a wrong route, the same way the pilot lesson's part (c) final marks are tied to the real numeric answer rather than to correct use of a wrong one.
A1

Named traps

ambiguous-case-obtuse-solution-missed: The one verified, quoted trap this entire lesson is anchored to. A real WMA11 examiner report states directly: "The majority used the sine rule and achieved the correct acute angle. However, many failed to find the required obtuse angle." (Jan 2019, Q7). The mechanism is structural, not carelessness: $\sin\theta = \sin(180°-\theta)$ for every $\theta$ in $(0°,180°)$ , so a calculator's $\sin^{-1}$ key can only ever hand back one of the two angles a real triangle might need. Whenever the sine rule is used to find an angle from an SSA setup, checking $180°$ minus the calculator's value — and deciding, from the question's own wording or an angle-sum check, which one the actual triangle needs — is not an optional extra step; it is the second half of the method.
cosine-rule-mistaken-for-having-an-ambiguous-case-too: Not a phrase the research bank quotes for this topic — named here as a mechanically-motivated risk, the natural overcorrection once a student has learned to distrust the sine rule's angle-finding results. It doesn't transfer: $\cos\theta$ is strictly decreasing across the whole of $(0°,180°)$ (its gradient, $-\sin\theta$ , is negative throughout that open interval), so $\cos^{-1}$ of any value in $(-1,1)$ returns exactly one angle there, with nothing left to check. A negative value partway through a cosine-rule calculation for an angle is the ordinary signature of an obtuse result, not a sign that a second candidate is hiding.
wrong-opposite-pair-used-in-the-sine-rule: Not a phrase the research bank quotes for this specific topic — a mechanically-motivated risk rather than a verified examiner-report finding. The sine rule's ratio only means what it claims to mean when each side is paired with the angle directly opposite it — $\frac{a}{\sin A}$ , not $\frac{a}{\sin B}$ . Mislabelling which side is opposite which angle (easy to do the moment a triangle is drawn without every vertex clearly marked) produces a ratio that is not the sine rule at all, however correctly the rest of the arithmetic is carried out from there.
area-formula-uses-a-non-included-angle: Not a phrase the research bank quotes for this specific topic — a mechanically-motivated risk, distinct from (though related in shape to) a separately VERIFIED finding for a different formula: a real examiner report on a genuine sector-AREA question (a different formula, spec 3.2, not 3.1) states "the most common error was to omit the ½ in the area formula." That citation is about the SECTOR area formula, ½r²θ, not the triangle area formula, ½ab sin C — it is named here only as an honest, clearly-scoped analogy, not as evidence for this different formula. The triangle-specific risk this item actually names is different in kind: $\frac12 ab\sin C$ only gives the right answer when $C$ is the angle physically trapped between the two sides $a$ and $b$ being multiplied — substituting any other angle in the triangle produces a number that looks exactly as plausible as the correct one, with no obvious tell on the page.
obtuse-cosine-rule-result-treated-as-an-arithmetic-error: Not a phrase the research bank quotes for this specific topic — a mechanically-motivated risk, the natural over-caution once a student has internalised that the ambiguous case exists somewhere in this spec point. A negative value appearing mid-calculation when rearranging the cosine rule for an angle ( $\cos A = \frac{b^2+c^2-a^2}{2bc}$ , with $a^2$ larger than $b^2+c^2$ ) is not a sign that a sign has been dropped somewhere upstream — it is exactly what an obtuse angle A looks like, since $\cos\theta$ is negative for every $\theta$ between 90° and 180°. Second-guessing a correctly negative intermediate value, rather than trusting the injectivity argument above, is its own way to lose time or introduce a real error while 'fixing' a step that was never broken.
sine-rule-angle-computed-in-the-wrong-calculator-mode: A SECOND real, verified examiner-report finding, from the exact same question as the ambiguous-case trap above (Jan 2019, Q7(a)) — re-extracted directly from the same PDF this lesson's central trap is anchored to, and previously absent from both this lesson and the facts bank's own entry for this question: "Some candidates worked in radians; candidates should be advised to work in the same angle measure which was degrees in this question (as shown by the 35° for the given angle)." A calculator left in the wrong angle mode does not produce an error — $\sin^{-1}$ returns a real, confident-looking number whichever mode it is in, since both degrees and radians are valid inputs to it. The fix is procedural, not conceptual: check the calculator's angle-mode setting BEFORE substituting into $\sin^{-1}$ or $\cos^{-1}$ , matching whichever unit the question itself uses — for every question on this WMA11 spec point, that unit is degrees.

Retrieval — with feedback on every choice

Question 1

3 marks

Triangle $ABC$ has $a = 9$ cm, $b = 10$ cm, $c = 15$ cm. A student finds $\cos C = \frac{a^2+b^2-c^2}{2ab} \approx -0.244$ and stops, saying they must have made a sign error since a cosine can't be negative here. Are they right?

Question 2

2 marks

Triangle $PQR$ has $PQ = 9$ cm, $PR = 6$ cm, and angle $PQR = 40°$ (opposite side $PR$ ). What can be said about the number of triangles this data describes, before doing any further arithmetic?

Question 3

3 marks

Triangle $PQR$ : $PQ = 12$ cm, $QR = 8$ cm, angle $QPR = 27°$ . Given that angle $PRQ$ is obtuse, find angle $PRQ$ (1 d.p.).

Beyond the spec

Spec 3.1 asks only that the cosine rule be known and used — WMA11-verified-facts.md's own scope note (§1) confirms P1 excludes proof as an examined skill entirely, so no WMA11 question will ever ask for this derivation to be reproduced. It is included because seeing where $a^2 = b^2+c^2-2bc\cos A$ actually comes from is what makes the earlier claim believable — that the formula 'reduces' to Pythagoras' theorem when $C=90°$ isn't a coincidence to memorise, it's visible directly in the derivation once C's role in it is seen.

Place the triangle so vertex $A$ sits at the angle being related to the two sides that meet there, and drop a perpendicular from $C$ to line $AB$ , meeting it at point $D$ (taking $A$ acute, so $D$ falls between $A$ and $B$ — the argument extends to an obtuse $A$ with signed lengths, but the acute case already shows the whole mechanism). In right-angled triangle $ACD$ : $AD = b\cos A$ (adjacent over hypotenuse) and $CD = b\sin A$ (opposite over hypotenuse), where $b = AC$ . Then $DB = AB - AD = c - b\cos A$ , where $c = AB$ . Triangle $CDB$ is also right-angled (at $D$ ), so by Pythagoras: $BC^2 = DB^2 + CD^2$ , i.e. $a^2 = (c-b\cos A)^2 + (b\sin A)^2$ . Expand: $a^2 = c^2 - 2bc\cos A + b^2\cos^2A + b^2\sin^2A = c^2 - 2bc\cos A + b^2(\cos^2A+\sin^2A)$ . Since $\cos^2A+\sin^2A=1$ for every angle $A$ (Pythagoras' theorem applied to the same right-angled triangle that defines sine and cosine in the first place), this collapses to $a^2 = b^2+c^2-2bc\cos A$ — the cosine rule, derived rather than quoted. Set $A=90°$ and watch it collapse further: $\cos 90° = 0$ , so the whole $-2bc\cos A$ term vanishes and $a^2=b^2+c^2$ is left standing alone — Pythagoras' theorem, recovered as the special case of the cosine rule where the "included angle" happens to be a right angle. The cosine rule was never a separate fact bolted onto Pythagoras; it is what Pythagoras' theorem becomes once the constraint of a right angle is lifted and any angle at all is allowed to sit between the two known sides.

Reference — not a study method, a lookup

Sine rule: a/sinA = b/sinB = c/sinC — finds a side (safe, ASA/AAS) or an angle (SSA — check the ambiguous case).
Cosine rule: a² = b²+c²−2bc cosA (SAS, finds a side) or cosA = (b²+c²−a²)/(2bc) (SSS, finds an angle) — NEVER ambiguous; cos is one-to-one on (0°,180°).
Area = ½ab sinC — C must be the angle physically INCLUDED between sides a and b, not any other angle in the triangle.
Ambiguous case ONLY when finding an ANGLE by the sine rule from SSA data: found sinθ=k, check θ AND 180°−θ, then use the question's own wording or an angle-sum check to decide which (or both) is valid.
Formula booklet has the cosine rule only — sine rule, area formula, arc length and sector area all have to be memorised.

Not affiliated with or endorsed by Pearson Edexcel. This lesson's ambiguous-case content is anchored to TWO verified examiner-report findings, both from the SAME series and the SAME question — WMA11_01_1901_ER, Jan 2019 Q7(a), re-confirmed directly against the primary-source PDF by a bullet-coverage audit pass: "The majority used the sine rule and achieved the correct acute angle. However, many failed to find the required obtuse angle." (reused verbatim across the diagram, the central marked-solution, one trap-taxonomy item and one MCQ), and "Some candidates worked in radians; candidates should be advised to work in the same angle measure which was degrees in this question" (a second, independent trap from the same examiner-report paragraph, added by that same audit pass and reused in one trap-taxonomy item). That audit pass also re-verified the VERBATIM M/A/B structure of Q7(a) itself against the real mark-scheme PDF (M1, A1, A1 — the middle mark earned by EITHER candidate angle on its own, not specifically by finding both) and corrected this lesson's marked-solution and commonWrongPath, which had previously mislabelled that middle mark as a dependent METHOD mark ("dM1") requiring both candidates, a mark-type and dependency-structure error a real mark scheme does not support. No other examiner-report material exists in the research bank for the sine rule, the cosine rule, or the area formula specifically — this is thinner evidence than every other WMA11 lesson in this unit's original six, and thinner even than splitting-fractions-before-integrating.ts's single-series topic, which at least had a second, four-series-confirmed trap to draw on. Three of the six trap-taxonomy items explicitly say so in their own text (cosine-rule-mistaken-for-having-an-ambiguous-case-too, wrong-opposite-pair-used-in-the-sine-rule, obtuse-cosine-rule-result-treated-as-an-arithmetic-error) — these are mechanically-motivated risks named for completeness, not phrases the research bank quotes for this topic. The area-formula trap item is handled with a deliberate, visible caveat of its own kind: it names a REAL verified finding — "the most common error was to omit the ½ in the area formula" — but that finding is about the SECTOR area formula (½r²θ, spec 3.2, verified in sector-area-and-arc-length.ts), not the TRIANGLE area formula (½ab sinC, spec 3.1) this lesson teaches, and the item says so explicitly rather than letting the citation appear to transfer. Everything else is either the spec's own verbatim content (§1, directly transcribed), the general marking-guidance conventions verbatim-verified in §3 and already reused by every WMA11 sibling lesson (the "quote the formula first" advice, the M/A/B mark-type definitions, the ft/cao conventions), or VERIDIAN-original: every specific triangle, side length, angle and mark allocation in this lesson's worked-chain, chain-drill, marked-solution and MCQ blocks (the central AB=10/BC=7/angle-A=35° triangle included) was computed and independently cross-checked with a calculator before being written in, not transcribed from any real WMA11 question, because Q7's own numbers were kept out of this lesson's worked examples deliberately, to keep the citation scoped to the finding rather than the question's specific figures.

Question 13 marks

Triangle $ABC$ has $a = 9$ cm, $b = 10$ cm, $c = 15$ cm. A student finds $\cos C = \frac{a^2+b^2-c^2}{2ab} \approx -0.244$ and stops, saying they must have made a sign error since a cosine can't be negative here. Are they right?

No — $\cos C \approx -0.244$ is correct, and it means angle $C$ (opposite the longest side, $c=15$ ) is obtuse: $C \approx 104.1°$ (1 d.p.).
Correct. $c=15$ is the longest side, so $C$ is the largest angle in the triangle — large enough here to be obtuse, and $\cos\theta$ is genuinely negative for every obtuse $\theta$ in $(90°,180°)$ . There is exactly one angle in $(0°,180°)$ with this cosine value, found directly by $\cos^{-1}(-0.244)$ , with nothing further to check.
BYes — cosine values in a valid triangle calculation should always come out positive, so a negative result means the formula was substituted wrong
This is exactly the mechanically-motivated trap the negative-value-treated-as-an-error item names. Cosine is genuinely negative for every angle between 90° and 180°, and a triangle's largest angle, opposite its longest side, is very often obtuse — nothing about a negative value here signals a mistake.
CThe student is right to be suspicious, but for the wrong reason — the real issue is that this triangle needs a second, acute-angled solution checked as well
The cosine rule for an angle never has a second solution to check — that is specifically a sine-rule phenomenon, arising because $\sin\theta$ (unlike $\cos\theta$ ) takes each value twice on $(0°,180°)$ . There is nothing missing here; $C \approx 104.1°$ is already the complete, unique answer.
DIt can't be determined without knowing whether the triangle is acute or obtuse in advance
That is exactly what this calculation is finding OUT, not something that needs to be known beforehand. The cosine rule works identically regardless of the triangle's shape — the sign of the result is the answer to "acute or obtuse," not an input to it.

Traps tested: Obtuse cosine rule result treated as an arithmetic error · Cosine rule mistaken for having an ambiguous case too · Shape of triangle assumed as a prerequisite

Question 22 marks

Triangle $PQR$ has $PQ = 9$ cm, $PR = 6$ cm, and angle $PQR = 40°$ (opposite side $PR$ ). What can be said about the number of triangles this data describes, before doing any further arithmetic?

It might be none, one, or two — this needs checking, since side $PQ$ (opposite the unknown angle) is longer than side $PR$ (opposite the given angle), and the given angle is acute.
Correct, and this is the honest answer before computing anything: an acute given angle with the OTHER given side (here PQ=9, opposite the unknown angle R) LONGER than the side opposite the KNOWN angle (PR=6) is exactly the ambiguous-case shape — and in this specific case (checking $\sin R = \frac{9\sin 40°}{6} \approx 0.9642$ ) it does resolve to two valid triangles, $R \approx 74.6°$ or $R \approx 105.4°$ , both giving an angle sum under $180°$ once added to the given $40°$ — but that has to be checked, not assumed from the shape of the data alone.
BExactly two, automatically, since this is an SSA (two-sides-and-a-non-included-angle) setup
SSA is the shape where two solutions are POSSIBLE, not guaranteed. Some genuine SSA triangles resolve to exactly one valid solution (the algebra returns two candidate angles, but one makes the angle sum exceed 180° and is thrown out), and some resolve to none at all (the required sine value comes out greater than 1). Assuming two without checking is the same class of error as assuming one without checking — both skip the actual verification step.
CExactly one, since a triangle's shape is always fully determined by two sides and any angle
Two sides and a non-included angle is specifically the one combination that does NOT always determine a unique triangle — that is the entire content of spec 3.1 naming an ambiguous case at all. ASA, AAS and SAS data each do determine a unique triangle; SSA is the exception.
DIt cannot be a valid triangle at all, since the given angle is not between the two given sides
A non-included angle is unusual to be GIVEN, but it doesn't make the triangle invalid — it makes the sine rule the right tool and raises exactly the question this item is testing: how many triangles satisfy this data, which needs checking rather than ruling out.

Traps tested: Ambiguous case assumed to always give two solutions · Ssa treated as equivalent to the unambiguous cases · Non included given angle assumed invalid

Question 33 marks

Triangle $PQR$ : $PQ = 12$ cm, $QR = 8$ cm, angle $QPR = 27°$ . Given that angle $PRQ$ is obtuse, find angle $PRQ$ (1 d.p.).

$137.1°$
Correct. $\frac{\sin R}{PQ} = \frac{\sin P}{QR}$ gives $\sin R = \frac{12\sin 27°}{8} \approx 0.6810$ , so $R \approx 42.9°$ or $R \approx 180°-42.9°=137.1°$ . The question states angle PRQ is obtuse, so the second candidate is the one required — exactly the check the Jan 2019 Q7 finding describes many candidates skipping.
B $42.9°$
This is the correct CALCULATOR value and the wrong FINAL answer — $\sin^{-1}(0.6810)$ returns the acute candidate directly, but the question's own stated condition (angle PRQ is obtuse) rules this one out. This is exactly the pattern a real examiner report names: "achieved the correct acute angle. However, many failed to find the required obtuse angle."
CThere is not enough information to find angle PRQ uniquely
There is enough information — the sine rule gives two mathematically possible values, and the question's own stated condition (obtuse) picks out exactly one of them. The information isn't missing; the second step (checking the supplementary angle, then applying the given condition) is what's needed.
D $110.3°$
This doesn't match either of the two angles the sine rule actually produces here ( $42.9°$ or its supplement, $137.1°$ ) — a genuine arithmetic slip rather than a conceptual one. Recomputing $\frac{12\sin 27°}{8}$ carefully, and checking the result against $\sin^{-1}$ and its supplement specifically, is the fix.

Traps tested: Ambiguous case obtuse solution missed · Ambiguous case treated as unsolvable rather than two branched · Arithmetic slip in the sine rule substitution

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Examiner report: Jan 2019 · Q7 — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA11.

The Sine Rule, the Cosine Rule, and the Ambiguous Case

Four shapes of triangle information, and which rule actually fits each one

Where the area formula comes from, and what the exam formula booklet actually gives you

Sine rule for a missing side (ASA) — no ambiguity, because both angles are given directly

Where the second triangle actually comes from — a compass, not just an equation

Why the cosine rule can never do this to you — and why the sine rule can

Cosine rule for a missing side (SAS)

Complete the chain — cosine rule for a missing angle (SSS), from three sides and no calculator shortcut

Area of a triangle — the one formula where using the wrong angle produces a wrong but plausible-looking number

Deep dive

Practice this for real