Sometimes two genuinely different triangles fit the same measurements. The sine rule finds one of them and stays silent about the other, so you have to know when to look.
Today's hook, Give a surveyor two sides and a non-included angle and there may be two different triangles that fit. Both are correct. A calculator will only ever show you one.
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Recall, your gut answer first
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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Recall, your gut answer first
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Your calculator gives $\sin^{-1}(0.5) = 30^\circ$. But what is $\sin 150^\circ$? Work it out, then say what that means for an angle found using the sine rule.
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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Two angles share the same sine
Work through the core explanation before applying it.
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Two angles share the same sine
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For any acute angle $\theta$, $\sin\theta = \sin(180^\circ - \theta)$. So $\sin 30^\circ = \sin 150^\circ = 0.5$. The inverse sine on a calculator returns only the acute one, so whenever the sine rule gives you an angle you must ask whether its obtuse partner also fits.
$\sin\theta = \sin(180^\circ - \theta)$ Ambiguity is possible when you know two sides and a non-included angle
Only for SSA
The ambiguous case arises when you know two sides and an angle NOT between them. SAS and SSS never produce two triangles.
Check the angle sum
A second solution is only valid if the obtuse angle plus the known angle is still under $180^\circ$. If not, discard it.
The cosine rule is never ambiguous
If the given data lets you start with the cosine rule, do so and the problem disappears.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
$\sin\theta = \sin(180^\circ - \theta)$, so two angles in a triangle can share a sine.
Ambiguity can arise only in the SSA case, two sides and a non-included angle.
A second solution is valid only if the two known angles still sum to less than $180^\circ$.
The cosine rule always returns a unique angle, so it is never ambiguous.
Understand
Concepts
Why the inverse sine hides a second solution while the inverse cosine does not.
Why the ambiguity is a property of the given data, not of the calculation.
Why checking the angle sum is what settles whether two triangles genuinely exist.
Can do
Skills
Recognise SSA data and test for a second solution.
Use geometric construction or a graphing application to see both triangles.
Choose the appropriate rule from the given information and justify the choice.
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Key terms
Ambiguous caseA situation where the given measurements fit two different triangles. Like this: $a = 7$, $b = 9$, $A = 40^\circ$ gives both $B \approx 55^\circ$ and $B \approx 125^\circ$.
SSAKnowing two sides and an angle that is not between them, the only arrangement that can be ambiguous. Like this: sides $a$ and $b$ with angle $A$, where $A$ faces $a$ rather than sitting between them.
Supplementary anglesTwo angles adding to $180^\circ$, which always have the same sine. Like this: $40^\circ$ and $140^\circ$ are supplementary, and both have sine $0.643$.
Inverse sineThe calculator operation that undoes a sine, returning only the acute answer. Like this: $\sin^{-1}(0.5)$ gives $30^\circ$, never $150^\circ$, even though both are correct.
Valid solutionA candidate angle that still leaves a positive third angle. Like this: if $A = 40^\circ$ and $B = 125^\circ$ then $C = 15^\circ$, so it is valid; if $B$ were $145^\circ$ the sum would exceed $180^\circ$.
Geometric constructionDrawing the triangle with compasses to see how many triangles the data allows. Like this: swinging an arc of radius $a$ from $C$ may cut the base line twice, once, or not at all.
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Why the inverse sine hides a solution
Work through the core explanation before applying it.
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Why the inverse sine hides a solution
core concept
On the unit circle, sine is the $y$-coordinate, and two different angles in $0^\circ$ to $180^\circ$ share the same height. Those two angles are $\theta$ and $180^\circ - \theta$.
A calculator must return one value, so it returns the acute one. When you write $B = \sin^{-1}(0.643) = 40^\circ$, the value $140^\circ$ has the same sine and may also fit.
The inverse cosine has no such problem, because cosine takes each value exactly once between $0^\circ$ and $180^\circ$: positive for acute angles, negative for obtuse ones.
This is why the ambiguity belongs to the sine rule alone. It is not a flaw in the sine rule; it is a property of the sine function on the interval a triangle angle lives in.
Quick check: if $\sin B = 0.5$ and $B$ is an angle in a triangle, what are the possible values of $B$?
$\sin\theta = \sin(180^\circ - \theta)$, so two triangle angles can share a sine. The inverse sine returns only the acute one. The inverse cosine is unique because cosine changes sign at $90^\circ$, which is why only the sine rule can be ambiguous.
Pause, copy the identity $\sin\theta = \sin(180^\circ-\theta)$, an example such as $\sin 40^\circ = \sin 140^\circ$, and the reason cosine avoids the problem, into your book.
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Testing whether a second triangle exists
core concept
We just saw that every sine rule angle has a supplementary partner with the same sine. That raises a question: does that partner always give a real second triangle? This card answers it → only if the angles still fit inside $180^\circ$, which is a one-line check.
Work out the acute angle from the sine rule, then form its supplement by subtracting from $180^\circ$.
Add the supplement to the angle you were originally given. If the total is less than $180^\circ$ there is room for a third angle, so a second triangle exists and both answers are correct.
If the total is $180^\circ$ or more, the second triangle is impossible and you discard it. Example: $A = 40^\circ$, $\sin B = 0.643$ gives $B = 40^\circ$ or $140^\circ$; since $40 + 140 = 180$, the obtuse case fails by exactly nothing and is rejected.
Construct it to see it. Draw the known angle and side, then swing an arc of the second known length from the far vertex. The arc may cross the base line twice, giving two triangles, once, or not at all. The syllabus names construction and graphing applications explicitly.
Fill the blank: if the sine rule gives $B = 35^\circ$, its supplementary partner is $180 - 35 = $ $^\circ$.
Find the acute angle, subtract from $180^\circ$ for its partner, then add the partner to the originally given angle. Under $180^\circ$ means two valid triangles; $180^\circ$ or more means the second is impossible. Constructing with an arc shows the same thing visually.
Pause, copy the two-step test (supplement, then angle-sum check) with a worked accept and a worked reject, and the arc construction, into your book.
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Choosing a rule, and avoiding the ambiguity
core concept
We just saw how to test whether the ambiguous case really produces two triangles. That raises a question: can you avoid the whole problem by picking a different rule? This card answers it → often yes, and the choice is decided entirely by what you were given.
If you know a complete side-and-opposite-angle pair, the sine rule is available and is usually quickest. Watch for ambiguity only when the unknown is an angle in an SSA arrangement.
If you know two sides and the included angle (SAS), or all three sides (SSS), use the cosine rule. Neither case is ever ambiguous.
A useful habit with SSA data: use the cosine rule to find the third side first, then the sine rule for the remaining angles. The cosine rule may give a quadratic in the unknown side, and two positive roots are exactly the two triangles, made visible rather than hidden.
Find the largest angle with the cosine rule. If you must find several angles, do the one opposite the longest side first with the cosine rule. Any obtuse angle must be that one, so every remaining angle is safely acute.
Which set of information can NOT produce two different triangles?
Sine rule when you have a side-and-opposite-angle pair; cosine rule for SAS or SSS, which are never ambiguous. With SSA, either test the supplement or find the third side by cosine rule first. Finding the largest angle first makes the rest safely acute.
Pause, copy the choose-a-rule triage (pair present means sine rule, SAS or SSS means cosine rule), and the largest-angle-first habit, into your book.
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · TWO VALID TRIANGLES
In triangle $ABC$, $a = 7$ cm, $b = 9$ cm and $A = 40^\circ$. Find all possible values of $B$.
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$\dfrac{\sin B}{9} = \dfrac{\sin 40^\circ}{7}$, so $\sin B = \dfrac{9\sin 40^\circ}{7} \approx 0.8264$
One triangle only, because SAS fixes it completely.
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Quick-fire practice
Work through the core explanation before applying it.
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Quick-fire practice
+10 XP
If $\sin B = 0.766$, give both possible values of $B$ to the nearest degree.
In triangle $ABC$, $a = 8$, $b = 10$, $A = 35^\circ$. Find both possible values of $B$.
In triangle $ABC$, $a = 20$, $b = 6$, $A = 80^\circ$. Show there is only one solution.
Which rule would you use given sides 5, 8 and the angle between them? Why?
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Revisit your inverse sine
Run the quick drill and copy the summary into your book.
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Revisit your inverse sine
At the start you found $\sin 150^\circ$. Explain in one sentence why a calculator cannot give you both answers, and describe the one check that decides whether the second one is a real triangle.
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
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Pick your answer, then rate your confidence, that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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Short answer
Write full responses, then check them against the model answers.
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Short answer
ApplyBand 54 marks
Q1. In triangle $ABC$, $a = 6$ cm, $b = 8$ cm and $A = 42^\circ$. Find all possible sizes of angle $B$, and justify whether each gives a valid triangle. (4 marks)
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ApplyBand 43 marks
Q2. In triangle $ABC$, $a = 25$, $b = 7$ and $A = 61^\circ$. Determine how many triangles are possible, showing your reasoning. (3 marks)
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UnderstandBand 43 marks
Q3. Explain why the ambiguous case can arise with the sine rule but never with the cosine rule, referring to the behaviour of sine and cosine between $0^\circ$ and $180^\circ$. (3 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: $50^\circ$ or $130^\circ$. Practice 2: $\sin B = \frac{10\sin 35^\circ}{8} \approx 0.717$, so $B \approx 45.8^\circ$ or $134.2^\circ$; both give an angle sum under $180^\circ$, so both are valid. Practice 3: $\sin B = \frac{6\sin 80^\circ}{20} \approx 0.295$, $B \approx 17.2^\circ$; the supplement $162.8^\circ$ gives $80 + 162.8 > 180$, so it is rejected. Practice 4: cosine rule, because the angle is included between the two known sides, which is SAS.
Q1 (4 marks): $\sin B = \frac{8\sin 42^\circ}{6} \approx 0.8921$ [1]. Acute: $B \approx 63.1^\circ$ [1]. Supplement: $B \approx 116.9^\circ$ [1]. $42 + 63.1 = 105.1 < 180$ and $42 + 116.9 = 158.9 < 180$, so both are valid and two triangles exist [1].
Q2 (3 marks): $\sin B = \frac{7\sin 61^\circ}{25} \approx 0.2448$ [1]. Acute: $B \approx 14.2^\circ$; supplement $\approx 165.8^\circ$ [1]. $61 + 165.8 = 226.8 > 180$, so the obtuse case is impossible and exactly one triangle exists [1].
Q3 (3 marks): Between $0^\circ$ and $180^\circ$ the sine function takes every value between 0 and 1 exactly twice, at $\theta$ and $180^\circ - \theta$ [1]. So $\sin^{-1}$ returns only the acute value and the obtuse partner stays hidden [1]. Cosine takes each value exactly once over the same interval, positive for acute and negative for obtuse angles, so $\cos^{-1}$ returns a unique angle and no ambiguity is possible [1].
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Review and finish
Take the module quiz if you are ready, then mark the lesson complete.
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Boss battle · Ambiguity Hunt
earn bronze · silver · gold
Decide how many triangles fit each set of measurements, and pick the rule that settles it fastest. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.