The Triangular Plot That Has Two Answers
A licensed surveyor in Kochi City is preparing a registration plan for a small triangular plot ABC. The boundary AC follows an existing road and measures 12.00 m. The angle at A between the road AC and the boundary AB was measured with a theodolite as 30.0°, and the opposite boundary BC was measured with a laser rangefinder as 8.00 m. The instruments are calibrated and the measurements are treated as exact. Because the plot borders a conservation zone, the local rule allows registration only if the area does not exceed 45 m²; a plan showing a larger area is rejected and must be resurveyed. The surveyor's assistant has drawn a single sketch with angle C clearly obtuse and has already sent it to the client. Before signing, the surveyor wants to know whether the three measurements determine the plot uniquely, what the alternatives are, and how the area rule bears on the choice.
1. Explain why the given data do not determine the triangle, and find both possible values of angle C, to one decimal place.
Read the worked answer
Write a = BC = 8 (opposite A), b = AC = 12, A = 30°. Two sides and a non-included angle are given, the ambiguous SSA case. By the sine rule,
sin B = b sin A / a = 12 × 0.5 / 8 = 0.75.
So B = 48.59° or B = 180° - 48.59° = 131.41°. Both must be checked for validity: with A = 30°, the obtuse value gives A + B = 161.41° < 180°, so a genuine second triangle exists. (Two solutions occur here precisely because b > a and sin B < 1.)
Then C = 180° - 30° - B:
- B = 48.59° gives C = 101.41° ≈ 101.4°;
- B = 131.41° gives C = 18.59° ≈ 18.6°.
Nothing in the raw measurements rules out either triangle, so the assistant's single sketch is unjustified.
2. Compute the area of both candidate plots and decide which one can be registered. Find a compact formula linking area to angle C first.
Read the worked answer
Use the two measured sides a = 8 and b = 12 with the included angle C:
Area = ½ ab sin C = ½ × 8 × 12 × sin C = 48 sin C.
- C = 101.41°: sin C = 0.9802, area = 47.05 m², which exceeds the 45 m² ceiling and is rejected.
- C = 18.59°: sin C = 0.3188, area = 15.30 m², which complies.
Corresponding third side: AB = c = a sin C / sin A = 16 sin C, giving 15.68 m in the obtuse case and 5.10 m in the acute case.
The registrable plot is therefore the acute-C triangle with C ≈ 18.6° and AB ≈ 5.1 m. The assistant's obtuse sketch is exactly the one that fails, and only by about 2 m² - a warning that the decision hinges on carrying the sine values accurately rather than on rounding.
3. Suppose the 45 m² ceiling is dropped and a neighbouring fence line instead fixes angle C at more than 90°. Which triangle is the plot, and what is its area? Then, keeping a = 8 and b = 12, determine the range of angle A for which the ambiguity exists at all.
Read the worked answer
C > 90° selects C = 101.41°, so B = 48.59°, and the area is 48 sin(101.41°) = 47.05 m², with AB = 16 sin(101.41°) = 15.68 m.
For the general question, sin B = 1.5 sin A. Two distinct triangles require: (1) 1.5 sin A < 1, otherwise no angle B exists; (2) the obtuse candidate B₂ = 180° - B₁ must satisfy A + B₂ < 180°, i.e. A < B₁. Since b = 12 > a = 8, the larger side is opposite the larger angle, so B₁ > A automatically whenever B₁ exists. Condition (2) costs nothing.
Hence the ambiguity exists exactly when sin A < 2/3, i.e. 0 < A < arcsin(2/3) ≈ 41.8°. At A = 41.8° the triangle is right-angled at B (B = 90°) and is unique; for larger A no triangle with these measurements exists. A surveyor who wants a unique plot should therefore increase the measured angle at A or measure the third side, not measure AB as well and hope.