The Ambiguous SSA Triangle in an Angle-Chase
In a geometry mock, triangle ABC has AB=10, AC=7, and ∠ABC=30°. A candidate applies the sine rule as 7/sin30°=10/sin C, obtains sin C=5/7, and takes C≈45.6°. They then find ∠A≈104.4° and use the law of cosines to compute BC≈13.56. Satisfied that all angles are positive and the side lengths look plausible, they state that the triangle is uniquely determined and do not consider any other value of C. A classmate objects that a supplementary angle for C might also work, but the candidate says the diagram shows an acute angle at C. No diagram is actually supplied, and the problem statement contains only the three given data. The teacher asks the class to decide how many non-congruent triangles exist and to compute every possible BC.
1. How many non-congruent triangles satisfy the data? Determine all possible lengths BC exactly.
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Let a=BC. Law of cosines at B: AC^2=a^2+AB^2−2·a·AB·cos30°, so 49=a^2+100−20a(√3/2)=a^2+100−10√3 a. Thus a^2−10√3 a+51=0. The discriminant is 300−204=96, so a=(10√3±√96)/2=5√3±2√6. Both are positive: 5√3−2√6≈3.76 and 5√3+2√6≈13.56. Both correspond to valid C because sin C=5/7 has two supplementary solutions and both leave A positive. Hence there are two non-congruent triangles.
2. Explain the flaw in the uniqueness claim and identify which BC corresponds to obtuse C.
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SSA can be ambiguous. The sine rule loses information because sin C=sin(180°−C). The candidate silently chose acute C only. The acute solution C≈45.6° gives A≈104.4°, so the side opposite A, namely BC, is the larger root 5√3+2√6≈13.56. The obtuse solution C≈134.4° gives A≈15.6°, so BC is the smaller root 5√3−2√6≈3.76. A diagram could suggest one case but cannot eliminate the other; the textual data alone permit both triangles.