Explain the central structure of law of sines, ambiguous case & applications using precise mathematical language.
Apply the governing relationship and verify results with an alternate representation or substitution.
State domain, unit, precision, and context restrictions whenever they affect the result.
Analyze a misconception and communicate a correction with evidence.
Essential question: How do identities and triangle laws transform expressions, solve periodic equations, and determine unknown measurements?
Governing relationship
a/sinA=b/sinB=c/sinC.
The Law of Sines solves AAS, ASA, and sometimes SSA triangles. SSA may yield zero, one, or two triangles, so the ambiguous case must be checked.
Worked examples
Always label lowercase side a opposite uppercase angle A, and similarly for B and C. Keep the calculator in degree mode for examples using °.
Investigation A · AAS triangle with A=40°, B=65°, a=10.
Find the third angle: C=180°−40°−65°=75°.
Use paired opposites: a/sin A=b/sin B, so b=10 sin65°/sin40°≈14.10.
Likewise c=10 sin75°/sin40°≈15.03.
Check that the longest side c is opposite the largest angle C. Verify b/sin65°≈a/sin40°.
Domain: The denominator sin40° is not zero. The given A+B=105° leaves a valid positive third angle.
Investigation B · Ambiguous SSA: A=30°, a=7, b=10.
Compute the perpendicular altitude h=b sin A=10 sin30°=5. Here 5<7<10, so two triangles may exist.
Law of Sines gives sin B=(b sin A)/a=5/7. The principal solution is B₁≈45.58°.
Sine also has the supplementary angle B₂=180°−B₁≈134.42°. Check both angle sums: C₁=180°−30°−45.58°≈104.42° and C₂=180°−30°−134.42°≈15.58°. Both are positive.
Compute c₁=7sin C₁/sin30°≈13.56 and c₂=7sin C₂/sin30°≈3.76. These are two different, valid triangles.
Verify both with a/sin A=b/sin B. The first triangle has a large c and the second a much smaller c; do not discard either without checking.
Why there are sometimes 0, 1 or 2: For the same acute A=30° and b=10: a=4 is below h=5 (zero); a=5 is tangent, giving B=90° (one); a=7 yields two; a=12 is at least b (one). For an obtuse A, side a must be the longest.
Visual model · side–angle correspondence
When the information is SSA, use the altitude and both possible inverse-sine angles; the sketch alone cannot rule out the second triangle.
Guided and independent practice
AAS: Given A=35°, B=70°, and a=8, compute C and side b. Check which side should be the largest.
SSA classification: with A=30° and b=10, classify a=4, a=5, a=7 and a=12 as zero, one or two triangles. Explain using altitude and triangle angle sums.
Explain why A=110°, a=8, b=8 has no nondegenerate solution, although the ratio sin B may be numerically valid.
For the two-solution A=30°, a=7, b=10 case, compute both C values before using the Law of Sines for c.
Open Triangle Law Expedition and check the independent result while keeping game practice separate from formal mastery.
Independent transfer · check only after solving
With A=30°, b=10 and a=5, find B and C. Why is the triangle unique?
With A=110°, b=8 and a=8, state the number of valid triangles and justify your answer.
In the A=40°, B=65°, a=10 example, what is the value of c to two decimals?
Reveal worked verification and answers
The altitude is h=10sin30°=5, exactly a. Hence B=90°, C=60° and there is one right triangle.
Zero. Because A is obtuse, a must be strictly longer than side b, but a=b. Equivalently sin B=sin110° gives B=70° or 110°, neither leaving a positive C.
c=10sin75°/sin40°≈15.03; C is the largest angle, so c must be the largest side.
Differentiated pathways
Foundation
Review prerequisite vocabulary, annotate each worked step, complete the first four prompts, and use a table or graph to support symbolic work.
Core
Complete every prompt, explain domain and verification, correct errors, and submit the applied task with clear mathematical communication.
Extended
Generalize the relationship, compare methods, use technology strategically, analyze sensitivity, and create a proof, counterexample, or advanced model.
Application laboratory
Solve a surveying or navigation triangle and document ambiguity.
Required evidence: defined quantities and units; at least two representations; transparent calculations; verification; one limitation; and a defensible conclusion.
Misconception and correction
Common misconception: Automatically choosing the acute inverse-sine angle and ignoring a possible obtuse solution.
Write the first invalid step, the governing rule that corrects it, and one counterexample or verification that demonstrates the correction.
Mastery evidence
Accurate symbolic work with preserved equivalence.
Correct domain, interval, unit, and precision statements.
At least one independent verification.
A correction explaining why an incorrect approach fails.
A clear application conclusion that distinguishes mathematical result from recommendation.
Reflection
What structure did you recognize? Which representation was most useful? What assumption or restriction mattered most? What will you do differently on the next problem?