Ambiguous Triangle from Survey Data and a Fencing Budget
A land surveyor measures triangle ABC and records AB = 8 m, BC = 6 m, and angle BAC = 30°. The surveyor needs the length AC to order fencing. Because the given angle is not between the two known sides, a colleague warns that the data may not determine a unique triangle. The surveyor computes sin C = 2/3 and obtains two possible angles for C that both satisfy the angle-sum condition. One possible triangle is wide and shallow; the other is narrow and tall. The fencing budget allows a perimeter of at most 20 m. The surveyor must list both possible values of AC to the nearest 0.1 m, decide which triangle, if either, fits the budget, and explain why using the law of cosines directly would require solving a quadratic with two roots. The survey is on level ground, and all lengths are in meters. The surveyor also must state whether the two solutions correspond to genuinely different triangles or to the same triangle labeled differently.
1. Use the law of sines to find both possible values of AC, to the nearest 0.1 m.
Read the worked answer
In triangle ABC, side BC = 6 is opposite angle A = 30°, and side AB = 8 is opposite angle C. By the law of sines: sin C / 8 = sin 30° / 6. Thus sin C = 8·(1/2)/6 = 2/3.
This gives two possible angles: C1 = arcsin(2/3) ≈ 41.81° and C2 = 180° − 41.81° = 138.19°. Both are possible because in each case A + C < 180°: 30° + 41.81° = 71.81°, and 30° + 138.19° = 168.19°.
Then B = 180° − 30° − C. Let b = AC, which is opposite B. Again by the law of sines, b / sin B = 6 / sin 30° = 12, so b = 12 sin B.
- For C1 = 41.81°, B1 = 108.19°, so b1 = 12 sin(108.19°) ≈ 12(0.9500) ≈ 11.4 m.
- For C2 = 138.19°, B2 = 11.81°, so b2 = 12 sin(11.81°) ≈ 12(0.2047) ≈ 2.5 m.
The two possible lengths are about 11.4 m and 2.5 m.
2. Which possible triangle, if either, fits a perimeter budget of at most 20 m? Are the two solutions genuinely different triangles?
Read the worked answer
The two perimeters are:
- Wide triangle: 8 + 6 + 11.4 = 25.4 m, which exceeds 20 m.
- Narrow triangle: 8 + 6 + 2.5 = 16.5 m, which fits the 20 m budget.
So only the narrow triangle fits the budget.
The two solutions are genuinely different triangles. They have different values of AC, different angles C and B, and different perimeters. They are not the same triangle relabeled: the given side AB = 8 and BC = 6 remain fixed, but the included angle at B changes substantially, so the shapes are noncongruent.
3. Set up the law of cosines directly and show that it produces a quadratic with the same two roots.
Read the worked answer
Let b = AC. Using the law of cosines at angle A, with BC = 6 opposite A and AB = 8 adjacent to A:
6^2 = 8^2 + b^2 − 2·8·b·cos 30°.
So 36 = 64 + b^2 − 16b(√3/2), which gives 36 = 64 + b^2 − 8√3 b. Rearranging: b^2 − 8√3 b + 28 = 0.
The discriminant is (8√3)^2 − 4·1·28 = 192 − 112 = 80. Thus b = [8√3 ± √80]/2 = 4√3 ± 2√5. Numerically, 4√3 ≈ 6.928 and 2√5 ≈ 4.472, giving b ≈ 11.40 or b ≈ 2.46. These match the two lengths found by the law of sines. Both roots are positive, which is why the data allow two valid triangles.