Sine-rule survey with two possible triangles
During a land survey, triangle ABC represents three boundary posts. The measured sides are AB=8.4 m and AC=6.1 m, and the measured angle at B is angle ABC=32.0 degrees. A junior surveyor uses the sine rule: sin C/8.4=sin 32/6.1, giving sin C≈0.7297, so C≈46.9 degrees. She then states that the remaining angle is A≈101.1 degrees and that side BC follows uniquely. A senior surveyor warns that the measurements are of the SSA type and that the inverse sine has a second solution in the range 0 to 180 degrees. The fence contractor needs every possible triangle consistent with the measurements, including the length BC, before ordering materials. The team must decide whether the junior's single answer is safe and how the ambiguous case should be resolved.
1. Find all possible values of angle ACB consistent with the data and determine the corresponding angle BAC and length BC for each. Explain why both are geometrically valid.
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By the sine rule, sin C/8.4=sin32/6.1, so sin C=8.4 sin32/6.1≈0.7297. The two angles in 0 to 180 degrees with this sine are C≈46.86 degrees and C≈133.14 degrees. If C≈46.86 degrees, then A=180-32-46.86≈101.14 degrees. The side BC is opposite A, so BC/sin A=6.1/sin32, giving BC≈11.51 sin101.14≈11.29 m. If C≈133.14 degrees, then A≈14.86 degrees, and BC≈11.51 sin14.86≈2.95 m. Both satisfy the triangle inequalities with sides 8.4 m and 6.1 m. Geometrically, the side AC=6.1 m can meet the ray from B in two positions because it is longer than the perpendicular height 8.4 sin32≈4.45 m but shorter than AB=8.4 m. Thus both triangles are valid.
2. The junior says 'the inverse sine button gives 46.9 degrees, so the obtuse answer is invalid because angles in a triangle must be acute.' Critique this reasoning and state a general condition for the SSA ambiguous case to produce two triangles.
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The statement that triangle angles must be acute is false; an obtuse triangle is still a triangle. The inverse sine function returns only the principal value in [-90,90], here 46.86 degrees, but the equation sin C=0.7297 has a second solution C=180-46.86=133.14 degrees in the triangle range. The SSA ambiguous case gives two triangles when the side opposite the given angle is shorter than the adjacent given side but longer than the perpendicular height. Here the side opposite the 32 degree angle is AC=6.1 m, the adjacent given side is AB=8.4 m, and the height is 8.4 sin32≈4.45 m. Since 4.45<6.1<8.4, two triangles exist. The junior should report both possible BC lengths, not just the acute-angle one.