Pure Mathematics 1

Deep Dive: Sine and Cosine Rules Ambiguous Case

Provenance: written 2026-09-29. Sources actually read for this document: research/veridian/WMA11-verified-facts.md in full for this topic — §1 (spec 3.1's verbatim wording), §2 (formula-booklet contents), §4.3 ("Sine/cosine rule, including the ambiguous case," Jan 2019 Q7 verbatim M/A/B breakdown and the Alt1(a) cosine-rule note), §7 (June 2022 Q2 flag), and the full "sine-and-cosine-rules-ambiguous-case.ts (spec 3.1)" lesson-audit-log entry; the live lesson file content/veridian/curriculum/wma11/sine-and-cosine-rules-ambiguous-case.ts (all 793 lines, read in full, specifically to avoid duplicating its teach blocks, worked chains, diagram, marked-solution, trap-taxonomy and MCQs); content/veridian/technique-hub/wma11/signal-topic-brief-sine-and-cosine-rules-ambiguous-case.md (in full); content/veridian/technique-hub/wma11/codex-formula-and-method-bank.md §1.3.1 and §2.3.1; and content/veridian/question-bank/wma11.ts (grepped — confirmed to contain no sine/cosine-rule or "ambiguous" content, so there is nothing there to cross-check against or avoid duplicating). No new primary-source PDF was fetched for this pass; every exam citation below is inherited from the facts bank's own already-verified extraction of WMA11_01_1901_QP/MS/ER, re-checked here only for correct attribution and word-count compliance on quotes. Every classification, worked micro-example and algebraic identity in §2–§4 below is original mathematical derivation, not an exam citation — each was checked numerically with Python (math.sin/asin/cos, exact float comparison) before being written in, the same discipline the live lesson's own header describes for its worked numbers. Not affiliated with or endorsed by Pearson Edexcel.

#CODEX Deep-Dive — Sine and Cosine Rules: The Ambiguous Case (WMA11, spec 3.1)

#0. What this document does not repeat

Spec 3.1 reads, verbatim: "The sine and cosine rules, and the area of a triangle in the form ½ab sin C — including the ambiguous case of the sine rule." The live lesson already builds, correctly and in full: which of the four data shapes (SSS, SAS, ASA/AAS, SSA) each rule actually fits; a from-scratch derivation of ½ab sin C; the exam formula booklet's real P1 contents (cosine rule only — sine rule and area formula must be memorised); a coordinate-geometry construction of where a second triangle geometrically comes from (a circle of radius a meeting a fixed ray at up to two points); a structural injectivity argument for why the cosine rule can never produce this ambiguity; four fully-worked chains/drills covering every rule in its safe form; a complete marked-solution built around the one verified examiner finding on this spec point; six trap-taxonomy items; and three MCQs. None of that is reproduced here.

What the lesson's single worked mechanism example does not do — because one numerical instance can't — is state the general rule for how many triangles a given SSA data set actually determines, or address the mark-scheme's own alternative route for solving SSA data. Both are genuine, spec-required content (the ambiguous case is, precisely, a solution-count question, and spec 3.1 names no single method as the required one). This document supplies both, plus an honest accounting of every real exam angle on record for this exact topic.

#1. Definitions, precisely

Opposite pair. A side and the angle facing it across the triangle — side a opposite angle A, and so on. The sine rule's ratio $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ only means what it claims when each fraction pairs a side with its own opposite angle.
SSA (two sides, non-included angle). Two given sides and a given angle that is not trapped between them — equivalently, the angle is opposite one of the two given sides, leaving the other given side "loose." This is the one data shape, among the four spec 3.1 recognises, where the sine rule's angle-finding step requires $\sin^{-1}$ of a ratio rather than a direct division.
The ambiguous case. The fact that $\sin\theta=\sin(180°-\theta)$ for every $\theta$ in the only range a triangle's angles can occupy, $(0°,180°)$ — so solving $\sin B = k$ by calculator returns one angle when the data may genuinely admit a second, geometrically distinct triangle with the supplementary angle in its place.
Candidate angle. Either of the two values, $\theta$ and $180°-\theta$ , that satisfy $\sin\theta=k$ on $(0°,180°)$ . Only some candidates survive as valid triangles — a candidate is valid exactly when it keeps the angle sum under $180°$ (equivalently, when it doesn't require the triangle's third angle to be negative or zero).

#2. The complete classification: how many triangles does a given SSA data set actually determine?

Fix the data as: angle $A$ , its opposite side $a$ , and the other given side $b$ (adjacent to $A$ , not opposite it — this is what makes the data SSA rather than SAS). Define $h = b\sin A$ — the height of the perpendicular dropped from the vertex opposite $b$ down to the ray carrying the unknown side, in the lesson's own compass-construction picture. This single number, together with a plain comparison of $a$ against $b$ , settles the whole question. Both branches below were checked exhaustively across a systematic numerical sweep (a range of $A$ values on either side of $90°$ , with $a$ swept finely through every named boundary) before being stated as fact.

Branch 1 — $A$ acute ( $0°<A<90°$ ):

Condition	Triangles	Why
$a<h$	0	$\sin B = h/a>1$ — no angle has a sine that large.
$a=h$	exactly 1	$\sin B=1$ , so $B=90°$ exactly — the ray is tangent to the circle of radius $a$ ; the two candidate positions from the lesson's own construction coincide.
$h<a<b$	exactly 2	Both candidates, $B_1$ and $180°-B_1$ , keep the angle sum under $180°$ — the circle crosses the ray at two genuinely distinct positive points.
$a=b$	exactly 1	The triangle is isosceles ( $A=B_1$ ); the supplementary candidate forces the third angle to exactly $0°$ — degenerate, not a real triangle.
$a>b$	exactly 1	Only the acute candidate keeps the angle sum under $180°$ ; the supplementary candidate always overshoots once $a$ exceeds $b$ .

Branch 2 — $A$ a right angle or obtuse ( $A\geq90°$ ): $h$ plays no role here at all — a systematic check across $A=100°,110°,115°,120°,135°,150°$ with $a$ swept through the region $h<a<b$ that produces two triangles in Branch 1 found zero valid triangles there in every case; the only boundary that matters is $a$ against $b$ directly:

Condition	Triangles	Why
$a\leq b$	0	Even where $\sin^{-1}$ still returns a value, both resulting candidates force the third angle to zero or negative — checked directly at $a=b$ : the only surviving candidate has $B=180°-A$ , giving third angle $180°-A-(180°-A)=0°$ exactly, a degenerate boundary, not a real triangle.
$a>b$	exactly 1	Only the acute candidate for $B$ survives the angle-sum check; the case never produces two.

Verified micro-examples (all five Branch-1 rows, one fixed pair $A=40°$ , $b=12$ , so $h=b\sin40°\approx7.7135$ ):

$a=6.71<h$ : $\sin B\approx1.150>1$ — no triangle.
$a=h\approx7.7135$ : $B=90°$ , third angle $=50°$ — exactly one, right-angled at $B$ .
$a=9.86$ ( $h<a<b$ ): $B\approx51.47°$ or $B\approx128.53°$ , giving third angles $\approx88.53°$ or $\approx11.47°$ — both valid, two triangles.
$a=12=b$ : $B=40°$ (isosceles, $A=B$ ), third angle $100°$ — exactly one; the supplementary candidate ( $B=140°$ ) would force a $0°$ third angle.
$a=14>b$ : $B\approx33.43°$ , third angle $\approx106.57°$ — exactly one; the supplementary candidate ( $B\approx146.57°$ ) overshoots $180°$ once added to $A$ .

Verified micro-examples, Branch 2 ( $A=115°$ , $b=12$ , so $h\approx10.876$ — shown to confirm it is irrelevant here):

$a=10\leq b$ : no triangle (checked at $a=10.9$ , $11.5$ and $11.9$ too — all inside the region that would give two triangles under Branch 1's rule, all give zero here).
$a=12=b$ : no triangle — the one surviving candidate is $B=65°$ , and $115°+65°=180°$ exactly, a degenerate (zero-area) case.
$a=15>b$ : $B\approx46.47°$ , third angle $\approx18.53°$ — exactly one.

The practical shortcut this classification actually earns. Since $h=b\sin A<b$ for every acute $A$ (with equality only in the limit $A\to90°$ ), the two-triangle window $h<a<b$ is always a subset of $a<b$ . So genuine ambiguity — a real fork between two valid triangles, not just two candidates from $\sin^{-1}$ — is possible only when the side opposite the given angle is strictly shorter than the other given side ( $a<b$ ), and even then only if $a$ clears the height threshold $h$ . If $a\geq b$ , or if the given angle $A$ is itself a right angle or obtuse, there is at most one valid triangle regardless of what $\sin^{-1}$ returns — the supplementary-angle check the lesson's central trap is built around is still the correct method to run (it costs nothing and catches the genuine cases), but its outcome is decided before any arithmetic starts, the moment the two given side lengths are compared.

#3. Where this connects back to the lesson's own construction

The lesson's mechanism block places vertex $A$ at the origin, side $AC=b$ at angle $A$ above the ray carrying $B$ , and finds $B$ as wherever the circle of radius $a$ centred on $C$ crosses that ray. Solving that intersection algebraically — $(t-b\cos A)^2+(b\sin A)^2=a^2$ — gives $t=b\cos A\pm\sqrt{a^2-h^2}$ , exactly the two candidate positions the lesson's own numerical instance ( $A=30°,a=5,b=9$ ) draws. The " $a<h$ gives none, $a=h$ gives one, $a>h$ gives two candidate crossing points" fact is just this square root turning imaginary, zero, or real — the same discriminant idea the lesson's sibling lessons use for a quadratic's roots, applied here to a circle-meets-line intersection instead. What the lesson's single instance doesn't show is that a candidate crossing point can still fail to count, because it lands on the negative side of the ray (behind $A$ , not a valid triangle vertex): the nearer crossing point, $t=b\cos A-\sqrt{a^2-h^2}$ , is positive exactly when $\sqrt{a^2-h^2}<b\cos A$ — which rearranges (squaring, since both sides are positive for acute $A$ ) to $a^2-b^2\sin^2A<b^2\cos^2A$ , i.e. $a^2<b^2(\sin^2A+\cos^2A)=b^2$ , i.e. $a<b$ . That is the same $a<b$ condition §2 reaches from the angle-sum argument — two independent routes to an identical boundary, which is exactly the kind of cross-check this course's own discipline expects before trusting a general claim.

#4. A second, mark-scheme-credited route through SSA data: the cosine rule as a quadratic

The facts bank's own record of the one fully-documented exam appearance of this topic (Jan 2019 Q7(a)) notes that the real mark scheme's Alt1(a) route solves the same SSA data with the cosine rule instead of the sine rule — forming a quadratic in the unknown third side, then applying the cosine rule a second time for the angle. The live lesson's own teach block states the sine rule is "the only tool that fits" SSA data; its closing audit log flags this as a real, unresolved tension, left for "whichever pass next touches this lesson's teach block." This section is that pass, scoped narrowly to resolving it here rather than editing the lesson itself.

The method, in general. With $A$ , opposite side $a$ , and adjacent side $b$ known, and $c$ the unknown third side, the cosine rule $a^2=b^2+c^2-2bc\cos A$ rearranges into a quadratic in $c$ :

c^2-(2b\cos A)c+(b^2-a^2)=0.

By the quadratic formula, $c=b\cos A\pm\sqrt{b^2\cos^2A-b^2+a^2}=b\cos A\pm\sqrt{a^2-h^2}$ — the same expression §3 derived geometrically for the crossing point $t$ , confirming the two methods are the same construction seen algebraically instead of with coordinates. The discriminant is $4(a^2-h^2)$ , so it reproduces §2's own $a$ -vs- $h$ boundary directly: negative discriminant ( $a<h$ ) means no real $c$ , zero discriminant ( $a=h$ ) means one repeated root, positive discriminant ( $h<a$ ) means two real roots — of which only the positive ones are valid side lengths. By Vieta's formulas the two roots multiply to $b^2-a^2$ : when $a<b$ that product is positive, so both roots share the same sign as their sum ( $2b\cos A$ , positive for acute $A$ ) and both are valid — two triangles; when $a>b$ the product is negative, so exactly one root is positive and one is negative (an algebraically real but geometrically meaningless "negative length") — exactly one triangle. This is the identical $a$ -vs- $b$ test from §2, reached a third, independent way.

Verified micro-example ( $A=40°$ , $a=8$ opposite $A$ , $b=12$ — the same two-triangle regime as §2's row three): the sine rule gives $B\approx74.62°$ or $B\approx105.38°$ . The cosine-rule quadratic, $c^2-(24\cos40°)c+(144-64)=0$ , has discriminant $\approx18.01$ and roots $c\approx11.3145$ and $c\approx7.0706$ — both positive, so both are valid third sides. Substituting each back into $\cos A=\frac{b^2+c^2-a^2}{2bc}$ recovers $A=40.0000°$ exactly for both, confirming the two roots correspond to the same two triangles the sine rule found, reached without ever computing $\sin^{-1}$ of anything.

What this genuinely resolves. The sine rule remains the didactically-intended default for SSA data — it is shorter, and it is what the lesson correctly teaches as the standard route. But "the only tool that fits" overstates the real mark scheme, which credits the cosine-rule quadratic as an equally valid Alt1(a) method for the exact same data shape. The two methods are not just both "valid answers" in the loose sense every alternative method is — they are, as shown above, the same underlying algebra ( $a^2-h^2$ under a square root) arrived at from two different starting formulas, which is why they always agree on how many triangles exist and never disagree about which ones.

#5. Synthesis: every real exam angle this topic has actually been tested from

This course's own SIGNAL brief for this topic gives the honest count directly: the ambiguous case is confirmed in 2 of 9 series checked — one fully documented, one thinly. Both are real; neither should be over-generalised, and nothing below states a likelihood beyond that count.

Jan 2019 Q7(a) — fully documented. A roof-truss SSA question (3 marks, M1/A1/A1, no dependent marks — the only confirmed mark-tariff shape on record for this topic). The examiner report names two distinct real traps from the same paragraph: candidates who "achieved the correct acute angle" but "failed to find the required obtuse angle," and, separately, candidates who worked "in radians" when the question's own angle (35°) was in degrees. The first is this lesson's central trap; the second is its sixth trap-taxonomy item. Both are already fully taught in the live lesson and are not repeated here beyond this synthesis line. What §2 above adds to this specific appearance: the roof-truss context resolves the ambiguity by diagram/context-embedded validity — a candidate has to reason about which candidate angle the physical shape permits, not compute an $a$ -vs- $b$ comparison, because the question's own diagram (rather than a stated inequality) is what fixes it.

June 2022 Q2 — thinly documented. Recorded in the facts bank and the SIGNAL brief only as: a sine-rule ambiguous-case question resolved by a stated side-length comparison, found by a later, narrower content-gap audit rather than a full question-by-question read of that series. No mark tariff, wording, or M/A/B breakdown survives in the research pass for this one — inventing any of that here would be exactly the fabrication this course's discipline exists to prevent, so none is offered. What can honestly be added: a stated comparison of the two given side lengths is, per §2's own classification, a genuinely different — and in one respect more efficient — resolution route than the diagram-based one above. If a question states outright which of the two given sides is longer, that alone can settle whether ambiguity is even mathematically possible before any trigonometry is done: $a\geq b$ rules out a second triangle immediately (§2, Branch 1's last two rows), while $a<b$ leaves the question open pending the $h$ comparison. A real exam question resolving the ambiguous case this way is independent confirmation that the $a$ -vs- $b$ test in §2 is not merely a tidy theoretical shortcut — it is a genuine, examiner-tested resolution mechanism in its own right, distinct from reading a diagram.

What is not claimed. Seven of the nine series this course's research has touched for WMA11 were never checked question-by-question for this specific topic, so their silence is not evidence the ambiguous case didn't appear in them — only that it hasn't been found there yet. No grade boundary, no percentage of candidates, and no frequency finer than "2 of 9 series checked" is stated anywhere in this document, matching the one mark-tariff data point (3 marks, M1/A1/A1) actually on record. The Alt1(a) cosine-rule route in §4 is mark-scheme-verified for the one fully-documented appearance (Jan 2019 Q7) specifically — it is not separately confirmed for June 2022 Q2, where no method breakdown survives at all.

#6. Closing scope note

Everything in §2–§4 is ordinary, checkable trigonometry — a solution-count classification, a coordinate-geometry cross-check, and an algebraic identity connecting two textbook methods — verified here by direct numerical computation across a systematic sweep of cases, exactly the way the live lesson's own header describes verifying its worked numbers. None of it is an empirical claim about how WMA11 candidates actually perform, and none of it should be read as such. The only empirical claims in this document are in §5, and each is traced to the facts bank's own two named findings (Jan 2019 Q7(a), fully documented; June 2022 Q2, thinly documented) — nothing beyond what those two entries actually support.