The Drone Survey of a Triangular Plot
A land surveyor maps a triangular plot ABC. From a benchmark, she finds AB = 120 m, and the angles at A and B are ∠A = 35° and ∠B = 62°. She needs the length of side AC and the area of the plot to file a permit. A junior surveyor proposes using the Law of Cosines with sides AB and angle B to find AC directly. The senior surveyor says that will not work as set up and instead uses the Law of Sines. Later, a second plot DEF is measured with DE = 80 m, EF = 95 m, and ∠D = 40°, and the crew must decide how many triangles fit those measurements before committing to a boundary.
1. Find AC and the area of triangle ABC. Show which law applies and why the junior surveyor's plan fails.
Read the worked answer
First find the third angle: ∠C = 180° − 35° − 62° = 83°.
Law of Sines: AC / sin B = AB / sin C. AC = AB · sin B / sin C = 120 · sin 62° / sin 83°. sin 62° ≈ 0.8829, sin 83° ≈ 0.9925. AC ≈ 120 · 0.8829 / 0.9925 ≈ 120 · 0.8896 ≈ 106.7 m.
Area = ½ · AB · AC · sin A = ½ · 120 · 106.7 · sin 35°. sin 35° ≈ 0.5736. Area ≈ 0.5 · 120 · 106.7 · 0.5736 ≈ 60 · 106.7 · 0.5736 ≈ 3672 m².
(Equivalently Area = ½ · AB · BC · sin B after finding BC; results agree.)
Why the junior plan fails: the Law of Cosines needs two sides and the INCLUDED angle, or all three sides. Here we know AB and ∠B, but ∠B is not included between two known sides for solving AC - the known side AB is adjacent to ∠B, and the side we want (AC) is opposite ∠B, but we lack BC. The Law of Cosines for AC would require BC, which is unknown. So the junior surveyor's setup is underdetermined; the Law of Sines (AAS) is the correct tool.
2. For triangle DEF with DE = 80 m, EF = 95 m, and ∠D = 40°, determine how many distinct triangles satisfy the data. Explain the ambiguity.
Read the worked answer
This is the ambiguous SSA case. The known angle ∠D = 40° is opposite side EF = 95 m, and the adjacent known side is DE = 80 m.
Use the Law of Sines to find ∠F (opposite DE): sin F / DE = sin D / EF → sin F = DE · sin D / EF = 80 · sin 40° / 95. sin 40° ≈ 0.6428, so sin F ≈ 80 · 0.6428 / 95 ≈ 51.42 / 95 ≈ 0.5413.
F ≈ arcsin(0.5413) ≈ 32.8°, or F ≈ 180° − 32.8° = 147.2°.
Check both against ∠D = 40°:
- Case 1: F ≈ 32.8°, then E = 180° − 40° − 32.8° = 107.2° (valid).
- Case 2: F ≈ 147.2°, then E = 180° − 40° − 147.2° = −7.2° (invalid, negative angle).
So only ONE triangle fits. The reason: the side opposite the given angle (95 m) is longer than the adjacent side (80 m), so the SSA configuration is not ambiguous here - the longer side opposite the angle forces a unique solution. The crew can commit to a single boundary; no second triangle exists.