Ambiguous Triangular Flower Bed with a Fixed Budget
A landscaping class is laying out a triangular flower bed ABC on flat ground. They measure AB = 9.0 m and AC = 7.0 m, and the angle at B is 40°. Because the measurements form a side-side-angle pattern, the teacher warns that the triangle may not be uniquely determined. The class must order edging for side BC and topsoil for the whole bed. The budget allows at most 22 m of edging and at most 25 m² of topsoil. Two possible layouts are on the table: a 'wide' one with an obtuse angle at C and a 'narrow' one with an acute angle at C. A supplier's quote is valid only if the class can state a single perimeter and area. The teacher also wants the class to explain why the sine rule gives two answers here but only one when AC is lengthened while AB and angle B stay fixed.
1. Use the sine rule to find both possible values of ∠ACB and the corresponding lengths of BC.
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In triangle ABC, side AB = 9 is opposite angle C, and side AC = 7 is opposite angle B = 40°. The sine rule gives sin C / 9 = sin 40° / 7, so sin C = 9 sin 40° / 7 ≈ 9(0.64279)/7 ≈ 0.82644. Thus C ≈ 55.75° or C ≈ 180° − 55.75° = 124.25°. Both are admissible because each leaves a positive angle at A: A ≈ 84.25° or A ≈ 15.75°, and 40° + C < 180° in both cases. This is the ambiguous SSA case. For BC, again by the sine rule, BC / sin A = 7 / sin 40°, so BC = 7 sin A / sin 40°. For A ≈ 84.25°, BC ≈ 7(0.9950)/0.64279 ≈ 10.84 m. For A ≈ 15.75°, BC ≈ 7(0.2714)/0.64279 ≈ 2.96 m. So the two layouts have third sides of about 10.84 m and 2.96 m.
2. Compute the area of each layout and decide which layouts satisfy the 25 m² topsoil limit.
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Use area = ½(AB)(AC) sin A = ½(9)(7) sin A = 31.5 sin A. For the wide layout, A ≈ 84.25°, so sin A ≈ 0.9950 and area ≈ 31.5(0.9950) ≈ 31.34 m². For the narrow layout, A ≈ 15.75°, so sin A ≈ 0.2714 and area ≈ 31.5(0.2714) ≈ 8.55 m². The 25 m² topsoil limit therefore rules out the wide layout and allows the narrow one. A student who computed only the acute value of C would have found the narrow layout and might wrongly conclude that the problem is unique; the obtuse value is the one that breaches the budget. The judgement is to report both areas and then eliminate one by the stated constraint, rather than choosing the familiar-looking acute triangle by habit.
3. Compare perimeters against the 22 m edging budget and explain when the ambiguity disappears as AC increases.
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Wide layout: perimeter ≈ 9 + 7 + 10.84 = 26.84 m, which exceeds 22 m. Narrow layout: perimeter ≈ 9 + 7 + 2.96 = 18.96 m, which fits the budget. Combining both constraints, only the narrow layout with C ≈ 55.75°, A ≈ 15.75°, BC ≈ 2.96 m and area ≈ 8.55 m² can be ordered. For the ambiguity: sin C = 9 sin 40° / AC. As AC increases, sin C decreases, and the obtuse alternative C′ = 180° − C becomes invalid once C′ + 40° ≥ 180°, i.e. once C ≥ 140°. That happens when sin C ≤ sin 140° = sin 40°, which requires 9 sin 40° / AC ≤ sin 40°, i.e. AC ≥ 9. So with AB = 9 and angle B = 40°, any AC ≥ 9 gives exactly one triangle, while AC < 9 can give two.